Co-ordinate Geometry : Circle
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Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, frac{1}{2})$. Find the slope of the line.
Comments: First I write the family of lines passing through the given point i.e., $y - frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 sqrt{1 + m^2}$ but the solution of this equation is lengthy.
I am seeking a shorter solution.
analytic-geometry circle
asked Nov 30 '18 at 4:35
prashant sharmaprashant sharma
757
$endgroup$
add a comment |
$begingroup$
Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, frac{1}{2})$. Find the slope of the line.
Comments: First I write the family of lines passing through the given point i.e., $y - frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 sqrt{1 + m^2}$ but the solution of this equation is lengthy.
I am seeking a shorter solution.
analytic-geometry circle
asked Nov 30 '18 at 4:35
prashant sharmaprashant sharma
757
$endgroup$
$begingroup$
Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 5:39
add a comment |
$begingroup$
Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, frac{1}{2})$. Find the slope of the line.
Comments: First I write the family of lines passing through the given point i.e., $y - frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 sqrt{1 + m^2}$ but the solution of this equation is lengthy.
I am seeking a shorter solution.
analytic-geometry circle
asked Nov 30 '18 at 4:35
prashant sharmaprashant sharma
757
$endgroup$
Problem: Circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 8x + 11 = 0$ cut off equal intercepts on a line through the point $(-2, frac{1}{2})$. Find the slope of the line.
Comments: First I write the family of lines passing through the given point i.e., $y - frac{1}{2} = m (x + 2)$ then by using the pythagorean theorem and the hypothesis I got a equation in $m$ i.e., $| 2m - 7/2| - |2m + 1/2| = 15 sqrt{1 + m^2}$ but the solution of this equation is lengthy.
I am seeking a shorter solution.
analytic-geometry circle
analytic-geometry circle
asked Nov 30 '18 at 4:35
prashant sharmaprashant sharma
757
asked Nov 30 '18 at 4:35
prashant sharmaprashant sharma
757
asked Nov 30 '18 at 4:35
prashant sharmaprashant sharma
757
asked Nov 30 '18 at 4:35
prashant sharmaprashant sharma
757
asked Nov 30 '18 at 4:35
prashant sharmaprashant sharma
757
757
$begingroup$
Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 5:39
add a comment |
$begingroup$
Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 5:39
$begingroup$
Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 5:39
$begingroup$
Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 5:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Line $$2mx-2y+4m+1=0 tag{1}$$
First circle $$x^2+y^2=1$$
centre $O=(0,0)$, radius $r_1=1$
distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$
Second circle $$x^2+y^2-8x+11=0$$
centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$
distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$
Equating semi-chord length: $r^2-d^2$
begin{align}
r_1^2-d_1^2 &= r_2^2-d_2^2 \
1-frac{(4m+1)^2}{4(m^2+1)} &=
5-frac{(12m+1)^2}{4(m^2+1)} \
m &= frac{-1pm sqrt{29}}{14}
end{align}
Only the negative root gives intersections:
answered Nov 30 '18 at 6:46
Ng Chung TakNg Chung Tak
14.6k31334
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Line $$2mx-2y+4m+1=0 tag{1}$$
First circle $$x^2+y^2=1$$
centre $O=(0,0)$, radius $r_1=1$
distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$
Second circle $$x^2+y^2-8x+11=0$$
centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$
distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$
Equating semi-chord length: $r^2-d^2$
begin{align}
r_1^2-d_1^2 &= r_2^2-d_2^2 \
1-frac{(4m+1)^2}{4(m^2+1)} &=
5-frac{(12m+1)^2}{4(m^2+1)} \
m &= frac{-1pm sqrt{29}}{14}
end{align}
Only the negative root gives intersections:
answered Nov 30 '18 at 6:46
Ng Chung TakNg Chung Tak
14.6k31334
$endgroup$
add a comment |
$begingroup$
Line $$2mx-2y+4m+1=0 tag{1}$$
First circle $$x^2+y^2=1$$
centre $O=(0,0)$, radius $r_1=1$
distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$
Second circle $$x^2+y^2-8x+11=0$$
centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$
distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$
Equating semi-chord length: $r^2-d^2$
begin{align}
r_1^2-d_1^2 &= r_2^2-d_2^2 \
1-frac{(4m+1)^2}{4(m^2+1)} &=
5-frac{(12m+1)^2}{4(m^2+1)} \
m &= frac{-1pm sqrt{29}}{14}
end{align}
Only the negative root gives intersections:
answered Nov 30 '18 at 6:46
Ng Chung TakNg Chung Tak
14.6k31334
$endgroup$
add a comment |
$begingroup$
Line $$2mx-2y+4m+1=0 tag{1}$$
First circle $$x^2+y^2=1$$
centre $O=(0,0)$, radius $r_1=1$
distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$
Second circle $$x^2+y^2-8x+11=0$$
centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$
distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$
Equating semi-chord length: $r^2-d^2$
begin{align}
r_1^2-d_1^2 &= r_2^2-d_2^2 \
1-frac{(4m+1)^2}{4(m^2+1)} &=
5-frac{(12m+1)^2}{4(m^2+1)} \
m &= frac{-1pm sqrt{29}}{14}
end{align}
Only the negative root gives intersections:
answered Nov 30 '18 at 6:46
Ng Chung TakNg Chung Tak
14.6k31334
$endgroup$
Line $$2mx-2y+4m+1=0 tag{1}$$
First circle $$x^2+y^2=1$$
centre $O=(0,0)$, radius $r_1=1$
distance of $O$ from $(1)$: $$d_1=left| frac{4m+1}{2sqrt{m^2+1}} right|$$
Second circle $$x^2+y^2-8x+11=0$$
centre $P=(4,0)$, radius $r_2=sqrt{4^2-11}=sqrt{5}$
distance of $P$ from $(1)$: $$d_2=left| frac{12m+1}{2sqrt{m^2+1}} right|$$
Equating semi-chord length: $r^2-d^2$
begin{align}
r_1^2-d_1^2 &= r_2^2-d_2^2 \
1-frac{(4m+1)^2}{4(m^2+1)} &=
5-frac{(12m+1)^2}{4(m^2+1)} \
m &= frac{-1pm sqrt{29}}{14}
end{align}
Only the negative root gives intersections:
answered Nov 30 '18 at 6:46
Ng Chung TakNg Chung Tak
14.6k31334
edited Dec 2 '18 at 7:15
answered Nov 30 '18 at 6:46
Ng Chung TakNg Chung Tak
14.6k31334
answered Nov 30 '18 at 6:46
Ng Chung TakNg Chung Tak
14.6k31334
answered Nov 30 '18 at 6:46
Ng Chung TakNg Chung Tak
14.6k31334
14.6k31334
add a comment |
add a comment |
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$begingroup$
Check case when $2m$ is between $-infty,-1/2,0; 0,7/2; 7/2,infty$
$endgroup$
– lab bhattacharjee
Nov 30 '18 at 5:39