Locus of Midpoints of chords in a circle.
$begingroup$
This question is a Conics/Locus problem:
The circle $x^2+y^2=25$ cuts the y axis above the x axis at A. Find the locus of the midpoints of all chords of this circle that have A as one endpoint.
I’ve reasoned that the answer will be a circle with radius 2.5, but I don’t know how to mathematically prove it. I assume you use the midpoint formula?
conic-sections locus
edited Nov 25 '18 at 4:52
Key Flex
7,77451232
asked Nov 25 '18 at 4:47
M. WeateM. Weate
394
$endgroup$
add a comment |
$begingroup$
This question is a Conics/Locus problem:
The circle $x^2+y^2=25$ cuts the y axis above the x axis at A. Find the locus of the midpoints of all chords of this circle that have A as one endpoint.
I’ve reasoned that the answer will be a circle with radius 2.5, but I don’t know how to mathematically prove it. I assume you use the midpoint formula?
conic-sections locus
edited Nov 25 '18 at 4:52
Key Flex
7,77451232
asked Nov 25 '18 at 4:47
M. WeateM. Weate
394
$endgroup$
$begingroup$
You have taken wrong radius it should be 5
$endgroup$
– priyanka kumari
Nov 25 '18 at 6:46
$begingroup$
@priyankakumari OP is referring to the radius of the circle corresponding to the answer for the locus, which is indeed $5/2$ (see my solution).
$endgroup$
– Anurag A
Nov 25 '18 at 7:49
add a comment |
$begingroup$
This question is a Conics/Locus problem:
The circle $x^2+y^2=25$ cuts the y axis above the x axis at A. Find the locus of the midpoints of all chords of this circle that have A as one endpoint.
I’ve reasoned that the answer will be a circle with radius 2.5, but I don’t know how to mathematically prove it. I assume you use the midpoint formula?
conic-sections locus
edited Nov 25 '18 at 4:52
Key Flex
7,77451232
asked Nov 25 '18 at 4:47
M. WeateM. Weate
394
$endgroup$
This question is a Conics/Locus problem:
The circle $x^2+y^2=25$ cuts the y axis above the x axis at A. Find the locus of the midpoints of all chords of this circle that have A as one endpoint.
I’ve reasoned that the answer will be a circle with radius 2.5, but I don’t know how to mathematically prove it. I assume you use the midpoint formula?
conic-sections locus
conic-sections locus
edited Nov 25 '18 at 4:52
Key Flex
7,77451232
asked Nov 25 '18 at 4:47
M. WeateM. Weate
394
edited Nov 25 '18 at 4:52
Key Flex
7,77451232
asked Nov 25 '18 at 4:47
M. WeateM. Weate
394
edited Nov 25 '18 at 4:52
Key Flex
7,77451232
edited Nov 25 '18 at 4:52
Key Flex
7,77451232
edited Nov 25 '18 at 4:52
Key Flex
7,77451232
7,77451232
asked Nov 25 '18 at 4:47
M. WeateM. Weate
394
asked Nov 25 '18 at 4:47
M. WeateM. Weate
394
asked Nov 25 '18 at 4:47
M. WeateM. Weate
394
394
$begingroup$
You have taken wrong radius it should be 5
$endgroup$
– priyanka kumari
Nov 25 '18 at 6:46
$begingroup$
@priyankakumari OP is referring to the radius of the circle corresponding to the answer for the locus, which is indeed $5/2$ (see my solution).
$endgroup$
– Anurag A
Nov 25 '18 at 7:49
add a comment |
$begingroup$
You have taken wrong radius it should be 5
$endgroup$
– priyanka kumari
Nov 25 '18 at 6:46
$begingroup$
@priyankakumari OP is referring to the radius of the circle corresponding to the answer for the locus, which is indeed $5/2$ (see my solution).
$endgroup$
– Anurag A
Nov 25 '18 at 7:49
$begingroup$
You have taken wrong radius it should be 5
$endgroup$
– priyanka kumari
Nov 25 '18 at 6:46
$begingroup$
You have taken wrong radius it should be 5
$endgroup$
– priyanka kumari
Nov 25 '18 at 6:46
$begingroup$
@priyankakumari OP is referring to the radius of the circle corresponding to the answer for the locus, which is indeed $5/2$ (see my solution).
$endgroup$
– Anurag A
Nov 25 '18 at 7:49
$begingroup$
@priyankakumari OP is referring to the radius of the circle corresponding to the answer for the locus, which is indeed $5/2$ (see my solution).
$endgroup$
– Anurag A
Nov 25 '18 at 7:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The given circle $x^2+y^2=25$ has radius $5$ and center $(0,0)$. So based on your description $A$ has coordinates $(0,5)$.
Let $M(h,k)$ be the mid-point of the chord whose one end is at point $A$ and the other end at point $P$ on the circle. Then coordinates of $P$ will be given by $left(2h, 2k-5right)$.
Since $P$ lies on the circle, so it should satisfy the equation of the given circle. This means
begin{align*}
(2h)^2+(2k-5)^2&=25\
4h^2+4k^2-20k & =0\
h^2+k^2-5k&=0
end{align*}
So the locus of these midpoints is the circle given by $x^2+y^2-5y=0$. This is a circle with radius $frac{5}{2}$ and center at $left(0,frac{5}{2}right)$
answered Nov 25 '18 at 4:55
Anurag AAnurag A
25.8k12249
$endgroup$
add a comment |
$begingroup$
Construct green triangle $ OA = OB =R $ , $C$ is given as mid point of chord $ AB,$ so, the triangle $ OCA $ is right angled, being congruent to $OCB$. Now $R=5,$ Using polar coordinates and then Cartesian,
$$ dfrac{r}{sin theta } = R = dfrac{r^2}{r sin theta } = dfrac{x^2+y^2}{y} $$
$$ x^2+y^2 -y ,R =0 $$
which is the equation of red circle as locus of $C$.
answered Nov 25 '18 at 10:50
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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oldest
votes
$begingroup$
The given circle $x^2+y^2=25$ has radius $5$ and center $(0,0)$. So based on your description $A$ has coordinates $(0,5)$.
Let $M(h,k)$ be the mid-point of the chord whose one end is at point $A$ and the other end at point $P$ on the circle. Then coordinates of $P$ will be given by $left(2h, 2k-5right)$.
Since $P$ lies on the circle, so it should satisfy the equation of the given circle. This means
begin{align*}
(2h)^2+(2k-5)^2&=25\
4h^2+4k^2-20k & =0\
h^2+k^2-5k&=0
end{align*}
So the locus of these midpoints is the circle given by $x^2+y^2-5y=0$. This is a circle with radius $frac{5}{2}$ and center at $left(0,frac{5}{2}right)$
answered Nov 25 '18 at 4:55
Anurag AAnurag A
25.8k12249
$endgroup$
add a comment |
$begingroup$
The given circle $x^2+y^2=25$ has radius $5$ and center $(0,0)$. So based on your description $A$ has coordinates $(0,5)$.
Let $M(h,k)$ be the mid-point of the chord whose one end is at point $A$ and the other end at point $P$ on the circle. Then coordinates of $P$ will be given by $left(2h, 2k-5right)$.
Since $P$ lies on the circle, so it should satisfy the equation of the given circle. This means
begin{align*}
(2h)^2+(2k-5)^2&=25\
4h^2+4k^2-20k & =0\
h^2+k^2-5k&=0
end{align*}
So the locus of these midpoints is the circle given by $x^2+y^2-5y=0$. This is a circle with radius $frac{5}{2}$ and center at $left(0,frac{5}{2}right)$
answered Nov 25 '18 at 4:55
Anurag AAnurag A
25.8k12249
$endgroup$
add a comment |
$begingroup$
The given circle $x^2+y^2=25$ has radius $5$ and center $(0,0)$. So based on your description $A$ has coordinates $(0,5)$.
Let $M(h,k)$ be the mid-point of the chord whose one end is at point $A$ and the other end at point $P$ on the circle. Then coordinates of $P$ will be given by $left(2h, 2k-5right)$.
Since $P$ lies on the circle, so it should satisfy the equation of the given circle. This means
begin{align*}
(2h)^2+(2k-5)^2&=25\
4h^2+4k^2-20k & =0\
h^2+k^2-5k&=0
end{align*}
So the locus of these midpoints is the circle given by $x^2+y^2-5y=0$. This is a circle with radius $frac{5}{2}$ and center at $left(0,frac{5}{2}right)$
answered Nov 25 '18 at 4:55
Anurag AAnurag A
25.8k12249
$endgroup$
The given circle $x^2+y^2=25$ has radius $5$ and center $(0,0)$. So based on your description $A$ has coordinates $(0,5)$.
Let $M(h,k)$ be the mid-point of the chord whose one end is at point $A$ and the other end at point $P$ on the circle. Then coordinates of $P$ will be given by $left(2h, 2k-5right)$.
Since $P$ lies on the circle, so it should satisfy the equation of the given circle. This means
begin{align*}
(2h)^2+(2k-5)^2&=25\
4h^2+4k^2-20k & =0\
h^2+k^2-5k&=0
end{align*}
So the locus of these midpoints is the circle given by $x^2+y^2-5y=0$. This is a circle with radius $frac{5}{2}$ and center at $left(0,frac{5}{2}right)$
answered Nov 25 '18 at 4:55
Anurag AAnurag A
25.8k12249
answered Nov 25 '18 at 4:55
Anurag AAnurag A
25.8k12249
answered Nov 25 '18 at 4:55
Anurag AAnurag A
25.8k12249
answered Nov 25 '18 at 4:55
Anurag AAnurag A
25.8k12249
25.8k12249
add a comment |
add a comment |
$begingroup$
Construct green triangle $ OA = OB =R $ , $C$ is given as mid point of chord $ AB,$ so, the triangle $ OCA $ is right angled, being congruent to $OCB$. Now $R=5,$ Using polar coordinates and then Cartesian,
$$ dfrac{r}{sin theta } = R = dfrac{r^2}{r sin theta } = dfrac{x^2+y^2}{y} $$
$$ x^2+y^2 -y ,R =0 $$
which is the equation of red circle as locus of $C$.
answered Nov 25 '18 at 10:50
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
$begingroup$
Construct green triangle $ OA = OB =R $ , $C$ is given as mid point of chord $ AB,$ so, the triangle $ OCA $ is right angled, being congruent to $OCB$. Now $R=5,$ Using polar coordinates and then Cartesian,
$$ dfrac{r}{sin theta } = R = dfrac{r^2}{r sin theta } = dfrac{x^2+y^2}{y} $$
$$ x^2+y^2 -y ,R =0 $$
which is the equation of red circle as locus of $C$.
answered Nov 25 '18 at 10:50
NarasimhamNarasimham
20.6k52158
$endgroup$
add a comment |
$begingroup$
Construct green triangle $ OA = OB =R $ , $C$ is given as mid point of chord $ AB,$ so, the triangle $ OCA $ is right angled, being congruent to $OCB$. Now $R=5,$ Using polar coordinates and then Cartesian,
$$ dfrac{r}{sin theta } = R = dfrac{r^2}{r sin theta } = dfrac{x^2+y^2}{y} $$
$$ x^2+y^2 -y ,R =0 $$
which is the equation of red circle as locus of $C$.
answered Nov 25 '18 at 10:50
NarasimhamNarasimham
20.6k52158
$endgroup$
Construct green triangle $ OA = OB =R $ , $C$ is given as mid point of chord $ AB,$ so, the triangle $ OCA $ is right angled, being congruent to $OCB$. Now $R=5,$ Using polar coordinates and then Cartesian,
$$ dfrac{r}{sin theta } = R = dfrac{r^2}{r sin theta } = dfrac{x^2+y^2}{y} $$
$$ x^2+y^2 -y ,R =0 $$
which is the equation of red circle as locus of $C$.
answered Nov 25 '18 at 10:50
NarasimhamNarasimham
20.6k52158
edited Nov 25 '18 at 10:56
answered Nov 25 '18 at 10:50
NarasimhamNarasimham
20.6k52158
answered Nov 25 '18 at 10:50
NarasimhamNarasimham
20.6k52158
answered Nov 25 '18 at 10:50
NarasimhamNarasimham
20.6k52158
20.6k52158
add a comment |
add a comment |
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$begingroup$
You have taken wrong radius it should be 5
$endgroup$
– priyanka kumari
Nov 25 '18 at 6:46
$begingroup$
@priyankakumari OP is referring to the radius of the circle corresponding to the answer for the locus, which is indeed $5/2$ (see my solution).
$endgroup$
– Anurag A
Nov 25 '18 at 7:49