Mixing
Find the points of intersection of the line $x + 16 = 7y$ and the circle $x^2 + y^2 - 4x + 2y = 20$
Clash Royale CLAN TAG#URR8PPP
up vote
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I have the following problem:
On a single set of coordinate axes, sketch the line $x + 16 = 7y$ and the circle $x^2 + y^2 - 4x + 2y = 20$ and find their points of intersection.
So far I have only the center of the circle and its radius, I already tried to eliminate $x$ to solve with $y$ without success, any clues?
Edit:
I already tried with $(7y - 16)^2 + y^2 - 4(7y - 16) + 2y - 20 = 0$ but the result is wrong, I checked what could be the mistake but I don't see it, maybe I skipped something with the algebra process.
algebra-precalculus analytic-geometry
edited Jul 26 at 2:48
Robert Soupe
9,97721947
asked Jul 25 at 23:05
nutshell_A
163
add a comment |Â
up vote
2
down vote
favorite
I have the following problem:
On a single set of coordinate axes, sketch the line $x + 16 = 7y$ and the circle $x^2 + y^2 - 4x + 2y = 20$ and find their points of intersection.
So far I have only the center of the circle and its radius, I already tried to eliminate $x$ to solve with $y$ without success, any clues?
Edit:
I already tried with $(7y - 16)^2 + y^2 - 4(7y - 16) + 2y - 20 = 0$ but the result is wrong, I checked what could be the mistake but I don't see it, maybe I skipped something with the algebra process.
algebra-precalculus analytic-geometry
edited Jul 26 at 2:48
Robert Soupe
9,97721947
asked Jul 25 at 23:05
nutshell_A
163
eliminate x to solve with yThat works.without successWhy, what happened? Please edit your question to include what you tried and how/where it failed.
– dxiv
Jul 25 at 23:08
You should obtain $$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
– gimusi
Jul 25 at 23:24
thanks to everyone!! seems to be that I skipped a number with the algebra (embarrasing -.-) but I was wondering if there's other way to solve it (without involving trigonometry)...
– nutshell_A
Jul 25 at 23:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following problem:
On a single set of coordinate axes, sketch the line $x + 16 = 7y$ and the circle $x^2 + y^2 - 4x + 2y = 20$ and find their points of intersection.
So far I have only the center of the circle and its radius, I already tried to eliminate $x$ to solve with $y$ without success, any clues?
Edit:
I already tried with $(7y - 16)^2 + y^2 - 4(7y - 16) + 2y - 20 = 0$ but the result is wrong, I checked what could be the mistake but I don't see it, maybe I skipped something with the algebra process.
algebra-precalculus analytic-geometry
edited Jul 26 at 2:48
Robert Soupe
9,97721947
asked Jul 25 at 23:05
nutshell_A
163
I have the following problem:
On a single set of coordinate axes, sketch the line $x + 16 = 7y$ and the circle $x^2 + y^2 - 4x + 2y = 20$ and find their points of intersection.
So far I have only the center of the circle and its radius, I already tried to eliminate $x$ to solve with $y$ without success, any clues?
Edit:
I already tried with $(7y - 16)^2 + y^2 - 4(7y - 16) + 2y - 20 = 0$ but the result is wrong, I checked what could be the mistake but I don't see it, maybe I skipped something with the algebra process.
algebra-precalculus analytic-geometry
edited Jul 26 at 2:48
Robert Soupe
9,97721947
asked Jul 25 at 23:05
nutshell_A
163
edited Jul 26 at 2:48
Robert Soupe
9,97721947
edited Jul 26 at 2:48
Robert Soupe
9,97721947
edited Jul 26 at 2:48
Robert Soupe
9,97721947
9,97721947
asked Jul 25 at 23:05
nutshell_A
163
asked Jul 25 at 23:05
nutshell_A
163
asked Jul 25 at 23:05
nutshell_A
163
163
eliminate x to solve with yThat works.without successWhy, what happened? Please edit your question to include what you tried and how/where it failed.
– dxiv
Jul 25 at 23:08
You should obtain $$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
– gimusi
Jul 25 at 23:24
thanks to everyone!! seems to be that I skipped a number with the algebra (embarrasing -.-) but I was wondering if there's other way to solve it (without involving trigonometry)...
– nutshell_A
Jul 25 at 23:56
add a comment |Â
eliminate x to solve with yThat works.without successWhy, what happened? Please edit your question to include what you tried and how/where it failed.
– dxiv
Jul 25 at 23:08
You should obtain $$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
– gimusi
Jul 25 at 23:24
thanks to everyone!! seems to be that I skipped a number with the algebra (embarrasing -.-) but I was wondering if there's other way to solve it (without involving trigonometry)...
– nutshell_A
Jul 25 at 23:56
eliminate x to solve with y That works. without success Why, what happened? Please edit your question to include what you tried and how/where it failed.– dxiv
Jul 25 at 23:08
eliminate x to solve with y That works. without success Why, what happened? Please edit your question to include what you tried and how/where it failed.– dxiv
Jul 25 at 23:08
You should obtain $$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
– gimusi
Jul 25 at 23:24
You should obtain $$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
– gimusi
Jul 25 at 23:24
thanks to everyone!! seems to be that I skipped a number with the algebra (embarrasing -.-) but I was wondering if there's other way to solve it (without involving trigonometry)...
– nutshell_A
Jul 25 at 23:56
thanks to everyone!! seems to be that I skipped a number with the algebra (embarrasing -.-) but I was wondering if there's other way to solve it (without involving trigonometry)...
– nutshell_A
Jul 25 at 23:56
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Solve for $x$ in the equation of the line:
$$x=7y-16$$
Then plug your new equation for $x$ into the equation of the circle:
$$left(7y-16right)^2+y^2-4left(7y-16right)+2y-20=y^2-5y+6=0$$
This is a quadratic in terms of $y$; the solutions to the equation are $y=2$ and $y=3$. Since you said that you couldn't find the correct answer using this formula, here is the process to solve this using the quadratic formula:
$$y=frac5pmsqrt25-4left(1right)left(6right)2$$
$$y=frac5pm12$$
$$y=2,y=3$$
You can plug these values back into the equation of the line to find the $x$-values at which the circle and line intersect; these will be $x=-2$ and $x=5$. Therefore, the points of intersection are $left(-2,2right)$ and $left(5,3right)$. We can check this graphically as well:
answered Jul 25 at 23:43
csch2
220211
add a comment |Â
up vote
1
down vote
Since you asked for another way to solve it, here’s a method for computing the intersection of a line and any conic. It’s overkill for a circle-line intersection, though.
The general equation $ax+by+c=0$ of a line can be written as $(a,b,c)^Tcdot(x,y,1)^T=0$, so a line can be represented by the homogeneous vector $mathbf l = [a,b,c]^T$, with every point $mathbf x$ on the line satisfying $mathbf l^Tmathbf x = 0$. Similarly, the equation of any conic can be written as $mathbf x^TCmathbf x=0$, where $C$ is a symmetric matrix. The matrix $[mathbf l]_times^TC[mathbf l]_times$, where $[mathbf l]_times$ is the “cross-product matrix†of $mathbf l$—the skew-symmetric matrix such that $[mathbf l]_timesmathbf x = mathbf ltimesmathbf x$—turns out to be a degenerate dual conic matrix that represents that point of intersection of the line and conic. That is, it is some scalar multiple of $mathbf pmathbf q^T+mathbf qmathbf p^T$, where $mathbf p$ and $mathbf q$ are the points of intersection. We “split†this dual conic by finding a value of $alpha$ such that $[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times$ is a rank-one matrix, from which we will be able to read $mathbf p$ and $mathbf q$ directly because it is a multiple of $mathbf pmathbf q^T$.
For your problem, $$mathbf l = beginbmatrix1\-7\16endbmatrix, [mathbf l]_times = beginbmatrix0&-16&-7\16&0&-1\7&1&0endbmatrix \
C = beginbmatrix1&0&-2\0&1&1\-2&1&-20endbmatrix \
[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times = beginbmatrix-500 & 100-16alpha & 75-7alpha \ 100+16alpha & 300 & 125-alpha \ 75+7alpha & 125+alpha & 50 endbmatrix.$$ This last matrix will have rank one if all of the $2times2$ minors vanish. Choosing the lower-right produces the quadratic equation $alpha^2-625=0$, so $alpha=pm25$. Choosing the negative root yields $$beginbmatrix-500&500&250\-300&300&150\-100&100&50endbmatrix.$$ We can take any row/column pair with a nonzero diagonal entry. Taking the last, we have $mathbf p = [250,150,50]^T$ and $mathbf q = [-100,100,50]^T$, which dehomogenize to $(5,3)$ and $(-2,2)$.
answered Jul 26 at 8:05
amd
25.8k2943
Where is he asking for other methods? Anyway your solution are always stimulating for more deep considerations. Bye
– gimusi
Jul 26 at 9:29
add a comment |Â
up vote
0
down vote
HINT
Yes we need to substitute $x$ (or $y$) from the line equation into the circle equation and solve the quadratic equation obtained in $y$ (or $x$), that is
$$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
answered Jul 25 at 23:09
gimusi
65k73583
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Solve for $x$ in the equation of the line:
$$x=7y-16$$
Then plug your new equation for $x$ into the equation of the circle:
$$left(7y-16right)^2+y^2-4left(7y-16right)+2y-20=y^2-5y+6=0$$
This is a quadratic in terms of $y$; the solutions to the equation are $y=2$ and $y=3$. Since you said that you couldn't find the correct answer using this formula, here is the process to solve this using the quadratic formula:
$$y=frac5pmsqrt25-4left(1right)left(6right)2$$
$$y=frac5pm12$$
$$y=2,y=3$$
You can plug these values back into the equation of the line to find the $x$-values at which the circle and line intersect; these will be $x=-2$ and $x=5$. Therefore, the points of intersection are $left(-2,2right)$ and $left(5,3right)$. We can check this graphically as well:
answered Jul 25 at 23:43
csch2
220211
add a comment |Â
up vote
2
down vote
accepted
Solve for $x$ in the equation of the line:
$$x=7y-16$$
Then plug your new equation for $x$ into the equation of the circle:
$$left(7y-16right)^2+y^2-4left(7y-16right)+2y-20=y^2-5y+6=0$$
This is a quadratic in terms of $y$; the solutions to the equation are $y=2$ and $y=3$. Since you said that you couldn't find the correct answer using this formula, here is the process to solve this using the quadratic formula:
$$y=frac5pmsqrt25-4left(1right)left(6right)2$$
$$y=frac5pm12$$
$$y=2,y=3$$
You can plug these values back into the equation of the line to find the $x$-values at which the circle and line intersect; these will be $x=-2$ and $x=5$. Therefore, the points of intersection are $left(-2,2right)$ and $left(5,3right)$. We can check this graphically as well:
answered Jul 25 at 23:43
csch2
220211
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Solve for $x$ in the equation of the line:
$$x=7y-16$$
Then plug your new equation for $x$ into the equation of the circle:
$$left(7y-16right)^2+y^2-4left(7y-16right)+2y-20=y^2-5y+6=0$$
This is a quadratic in terms of $y$; the solutions to the equation are $y=2$ and $y=3$. Since you said that you couldn't find the correct answer using this formula, here is the process to solve this using the quadratic formula:
$$y=frac5pmsqrt25-4left(1right)left(6right)2$$
$$y=frac5pm12$$
$$y=2,y=3$$
You can plug these values back into the equation of the line to find the $x$-values at which the circle and line intersect; these will be $x=-2$ and $x=5$. Therefore, the points of intersection are $left(-2,2right)$ and $left(5,3right)$. We can check this graphically as well:
answered Jul 25 at 23:43
csch2
220211
Solve for $x$ in the equation of the line:
$$x=7y-16$$
Then plug your new equation for $x$ into the equation of the circle:
$$left(7y-16right)^2+y^2-4left(7y-16right)+2y-20=y^2-5y+6=0$$
This is a quadratic in terms of $y$; the solutions to the equation are $y=2$ and $y=3$. Since you said that you couldn't find the correct answer using this formula, here is the process to solve this using the quadratic formula:
$$y=frac5pmsqrt25-4left(1right)left(6right)2$$
$$y=frac5pm12$$
$$y=2,y=3$$
You can plug these values back into the equation of the line to find the $x$-values at which the circle and line intersect; these will be $x=-2$ and $x=5$. Therefore, the points of intersection are $left(-2,2right)$ and $left(5,3right)$. We can check this graphically as well:
answered Jul 25 at 23:43
csch2
220211
answered Jul 25 at 23:43
csch2
220211
answered Jul 25 at 23:43
csch2
220211
answered Jul 25 at 23:43
csch2
220211
220211
add a comment |Â
add a comment |Â
up vote
1
down vote
Since you asked for another way to solve it, here’s a method for computing the intersection of a line and any conic. It’s overkill for a circle-line intersection, though.
The general equation $ax+by+c=0$ of a line can be written as $(a,b,c)^Tcdot(x,y,1)^T=0$, so a line can be represented by the homogeneous vector $mathbf l = [a,b,c]^T$, with every point $mathbf x$ on the line satisfying $mathbf l^Tmathbf x = 0$. Similarly, the equation of any conic can be written as $mathbf x^TCmathbf x=0$, where $C$ is a symmetric matrix. The matrix $[mathbf l]_times^TC[mathbf l]_times$, where $[mathbf l]_times$ is the “cross-product matrix†of $mathbf l$—the skew-symmetric matrix such that $[mathbf l]_timesmathbf x = mathbf ltimesmathbf x$—turns out to be a degenerate dual conic matrix that represents that point of intersection of the line and conic. That is, it is some scalar multiple of $mathbf pmathbf q^T+mathbf qmathbf p^T$, where $mathbf p$ and $mathbf q$ are the points of intersection. We “split†this dual conic by finding a value of $alpha$ such that $[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times$ is a rank-one matrix, from which we will be able to read $mathbf p$ and $mathbf q$ directly because it is a multiple of $mathbf pmathbf q^T$.
For your problem, $$mathbf l = beginbmatrix1\-7\16endbmatrix, [mathbf l]_times = beginbmatrix0&-16&-7\16&0&-1\7&1&0endbmatrix \
C = beginbmatrix1&0&-2\0&1&1\-2&1&-20endbmatrix \
[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times = beginbmatrix-500 & 100-16alpha & 75-7alpha \ 100+16alpha & 300 & 125-alpha \ 75+7alpha & 125+alpha & 50 endbmatrix.$$ This last matrix will have rank one if all of the $2times2$ minors vanish. Choosing the lower-right produces the quadratic equation $alpha^2-625=0$, so $alpha=pm25$. Choosing the negative root yields $$beginbmatrix-500&500&250\-300&300&150\-100&100&50endbmatrix.$$ We can take any row/column pair with a nonzero diagonal entry. Taking the last, we have $mathbf p = [250,150,50]^T$ and $mathbf q = [-100,100,50]^T$, which dehomogenize to $(5,3)$ and $(-2,2)$.
answered Jul 26 at 8:05
amd
25.8k2943
Where is he asking for other methods? Anyway your solution are always stimulating for more deep considerations. Bye
– gimusi
Jul 26 at 9:29
add a comment |Â
up vote
1
down vote
Since you asked for another way to solve it, here’s a method for computing the intersection of a line and any conic. It’s overkill for a circle-line intersection, though.
The general equation $ax+by+c=0$ of a line can be written as $(a,b,c)^Tcdot(x,y,1)^T=0$, so a line can be represented by the homogeneous vector $mathbf l = [a,b,c]^T$, with every point $mathbf x$ on the line satisfying $mathbf l^Tmathbf x = 0$. Similarly, the equation of any conic can be written as $mathbf x^TCmathbf x=0$, where $C$ is a symmetric matrix. The matrix $[mathbf l]_times^TC[mathbf l]_times$, where $[mathbf l]_times$ is the “cross-product matrix†of $mathbf l$—the skew-symmetric matrix such that $[mathbf l]_timesmathbf x = mathbf ltimesmathbf x$—turns out to be a degenerate dual conic matrix that represents that point of intersection of the line and conic. That is, it is some scalar multiple of $mathbf pmathbf q^T+mathbf qmathbf p^T$, where $mathbf p$ and $mathbf q$ are the points of intersection. We “split†this dual conic by finding a value of $alpha$ such that $[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times$ is a rank-one matrix, from which we will be able to read $mathbf p$ and $mathbf q$ directly because it is a multiple of $mathbf pmathbf q^T$.
For your problem, $$mathbf l = beginbmatrix1\-7\16endbmatrix, [mathbf l]_times = beginbmatrix0&-16&-7\16&0&-1\7&1&0endbmatrix \
C = beginbmatrix1&0&-2\0&1&1\-2&1&-20endbmatrix \
[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times = beginbmatrix-500 & 100-16alpha & 75-7alpha \ 100+16alpha & 300 & 125-alpha \ 75+7alpha & 125+alpha & 50 endbmatrix.$$ This last matrix will have rank one if all of the $2times2$ minors vanish. Choosing the lower-right produces the quadratic equation $alpha^2-625=0$, so $alpha=pm25$. Choosing the negative root yields $$beginbmatrix-500&500&250\-300&300&150\-100&100&50endbmatrix.$$ We can take any row/column pair with a nonzero diagonal entry. Taking the last, we have $mathbf p = [250,150,50]^T$ and $mathbf q = [-100,100,50]^T$, which dehomogenize to $(5,3)$ and $(-2,2)$.
answered Jul 26 at 8:05
amd
25.8k2943
Where is he asking for other methods? Anyway your solution are always stimulating for more deep considerations. Bye
– gimusi
Jul 26 at 9:29
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since you asked for another way to solve it, here’s a method for computing the intersection of a line and any conic. It’s overkill for a circle-line intersection, though.
The general equation $ax+by+c=0$ of a line can be written as $(a,b,c)^Tcdot(x,y,1)^T=0$, so a line can be represented by the homogeneous vector $mathbf l = [a,b,c]^T$, with every point $mathbf x$ on the line satisfying $mathbf l^Tmathbf x = 0$. Similarly, the equation of any conic can be written as $mathbf x^TCmathbf x=0$, where $C$ is a symmetric matrix. The matrix $[mathbf l]_times^TC[mathbf l]_times$, where $[mathbf l]_times$ is the “cross-product matrix†of $mathbf l$—the skew-symmetric matrix such that $[mathbf l]_timesmathbf x = mathbf ltimesmathbf x$—turns out to be a degenerate dual conic matrix that represents that point of intersection of the line and conic. That is, it is some scalar multiple of $mathbf pmathbf q^T+mathbf qmathbf p^T$, where $mathbf p$ and $mathbf q$ are the points of intersection. We “split†this dual conic by finding a value of $alpha$ such that $[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times$ is a rank-one matrix, from which we will be able to read $mathbf p$ and $mathbf q$ directly because it is a multiple of $mathbf pmathbf q^T$.
For your problem, $$mathbf l = beginbmatrix1\-7\16endbmatrix, [mathbf l]_times = beginbmatrix0&-16&-7\16&0&-1\7&1&0endbmatrix \
C = beginbmatrix1&0&-2\0&1&1\-2&1&-20endbmatrix \
[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times = beginbmatrix-500 & 100-16alpha & 75-7alpha \ 100+16alpha & 300 & 125-alpha \ 75+7alpha & 125+alpha & 50 endbmatrix.$$ This last matrix will have rank one if all of the $2times2$ minors vanish. Choosing the lower-right produces the quadratic equation $alpha^2-625=0$, so $alpha=pm25$. Choosing the negative root yields $$beginbmatrix-500&500&250\-300&300&150\-100&100&50endbmatrix.$$ We can take any row/column pair with a nonzero diagonal entry. Taking the last, we have $mathbf p = [250,150,50]^T$ and $mathbf q = [-100,100,50]^T$, which dehomogenize to $(5,3)$ and $(-2,2)$.
answered Jul 26 at 8:05
amd
25.8k2943
Since you asked for another way to solve it, here’s a method for computing the intersection of a line and any conic. It’s overkill for a circle-line intersection, though.
The general equation $ax+by+c=0$ of a line can be written as $(a,b,c)^Tcdot(x,y,1)^T=0$, so a line can be represented by the homogeneous vector $mathbf l = [a,b,c]^T$, with every point $mathbf x$ on the line satisfying $mathbf l^Tmathbf x = 0$. Similarly, the equation of any conic can be written as $mathbf x^TCmathbf x=0$, where $C$ is a symmetric matrix. The matrix $[mathbf l]_times^TC[mathbf l]_times$, where $[mathbf l]_times$ is the “cross-product matrix†of $mathbf l$—the skew-symmetric matrix such that $[mathbf l]_timesmathbf x = mathbf ltimesmathbf x$—turns out to be a degenerate dual conic matrix that represents that point of intersection of the line and conic. That is, it is some scalar multiple of $mathbf pmathbf q^T+mathbf qmathbf p^T$, where $mathbf p$ and $mathbf q$ are the points of intersection. We “split†this dual conic by finding a value of $alpha$ such that $[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times$ is a rank-one matrix, from which we will be able to read $mathbf p$ and $mathbf q$ directly because it is a multiple of $mathbf pmathbf q^T$.
For your problem, $$mathbf l = beginbmatrix1\-7\16endbmatrix, [mathbf l]_times = beginbmatrix0&-16&-7\16&0&-1\7&1&0endbmatrix \
C = beginbmatrix1&0&-2\0&1&1\-2&1&-20endbmatrix \
[mathbf l]_times^TC[mathbf l]_times+alpha[mathbf l]_times = beginbmatrix-500 & 100-16alpha & 75-7alpha \ 100+16alpha & 300 & 125-alpha \ 75+7alpha & 125+alpha & 50 endbmatrix.$$ This last matrix will have rank one if all of the $2times2$ minors vanish. Choosing the lower-right produces the quadratic equation $alpha^2-625=0$, so $alpha=pm25$. Choosing the negative root yields $$beginbmatrix-500&500&250\-300&300&150\-100&100&50endbmatrix.$$ We can take any row/column pair with a nonzero diagonal entry. Taking the last, we have $mathbf p = [250,150,50]^T$ and $mathbf q = [-100,100,50]^T$, which dehomogenize to $(5,3)$ and $(-2,2)$.
answered Jul 26 at 8:05
amd
25.8k2943
answered Jul 26 at 8:05
amd
25.8k2943
answered Jul 26 at 8:05
amd
25.8k2943
answered Jul 26 at 8:05
amd
25.8k2943
25.8k2943
Where is he asking for other methods? Anyway your solution are always stimulating for more deep considerations. Bye
– gimusi
Jul 26 at 9:29
add a comment |Â
Where is he asking for other methods? Anyway your solution are always stimulating for more deep considerations. Bye
– gimusi
Jul 26 at 9:29
Where is he asking for other methods? Anyway your solution are always stimulating for more deep considerations. Bye
– gimusi
Jul 26 at 9:29
Where is he asking for other methods? Anyway your solution are always stimulating for more deep considerations. Bye
– gimusi
Jul 26 at 9:29
add a comment |Â
up vote
0
down vote
HINT
Yes we need to substitute $x$ (or $y$) from the line equation into the circle equation and solve the quadratic equation obtained in $y$ (or $x$), that is
$$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
answered Jul 25 at 23:09
gimusi
65k73583
add a comment |Â
up vote
0
down vote
HINT
Yes we need to substitute $x$ (or $y$) from the line equation into the circle equation and solve the quadratic equation obtained in $y$ (or $x$), that is
$$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
answered Jul 25 at 23:09
gimusi
65k73583
add a comment |Â
up vote
0
down vote
up vote
0
down vote
HINT
Yes we need to substitute $x$ (or $y$) from the line equation into the circle equation and solve the quadratic equation obtained in $y$ (or $x$), that is
$$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
answered Jul 25 at 23:09
gimusi
65k73583
HINT
Yes we need to substitute $x$ (or $y$) from the line equation into the circle equation and solve the quadratic equation obtained in $y$ (or $x$), that is
$$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
answered Jul 25 at 23:09
gimusi
65k73583
answered Jul 25 at 23:09
gimusi
65k73583
answered Jul 25 at 23:09
gimusi
65k73583
answered Jul 25 at 23:09
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
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eliminate x to solve with yThat works.without successWhy, what happened? Please edit your question to include what you tried and how/where it failed.– dxiv
Jul 25 at 23:08
You should obtain $$(7y-16)^2+y^2-4(7y-16)+2y-20=0implies y^2-5y+6=0$$
– gimusi
Jul 25 at 23:24
thanks to everyone!! seems to be that I skipped a number with the algebra (embarrasing -.-) but I was wondering if there's other way to solve it (without involving trigonometry)...
– nutshell_A
Jul 25 at 23:56