Random point inside a circle
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A point inside a circle of radius $sqrt{50}$ lies $2$ units directly below a point on the circle, and $6$ units directly to the
right of a point on the circle. What is the distance from
the center of the circle to this point?
Can't find anything, this was rated as an easy question so i think there will be just one concept for this. I tried power of a point and and coordinate geometry but I cant answer it
geometry
edited Nov 16 at 9:27
MathFun123
516216
asked Nov 16 at 8:09
SuperMage1
852210
add a comment |
up vote
0
down vote
favorite
A point inside a circle of radius $sqrt{50}$ lies $2$ units directly below a point on the circle, and $6$ units directly to the
right of a point on the circle. What is the distance from
the center of the circle to this point?
Can't find anything, this was rated as an easy question so i think there will be just one concept for this. I tried power of a point and and coordinate geometry but I cant answer it
geometry
edited Nov 16 at 9:27
MathFun123
516216
asked Nov 16 at 8:09
SuperMage1
852210
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A point inside a circle of radius $sqrt{50}$ lies $2$ units directly below a point on the circle, and $6$ units directly to the
right of a point on the circle. What is the distance from
the center of the circle to this point?
Can't find anything, this was rated as an easy question so i think there will be just one concept for this. I tried power of a point and and coordinate geometry but I cant answer it
geometry
edited Nov 16 at 9:27
MathFun123
516216
asked Nov 16 at 8:09
SuperMage1
852210
A point inside a circle of radius $sqrt{50}$ lies $2$ units directly below a point on the circle, and $6$ units directly to the
right of a point on the circle. What is the distance from
the center of the circle to this point?
Can't find anything, this was rated as an easy question so i think there will be just one concept for this. I tried power of a point and and coordinate geometry but I cant answer it
geometry
geometry
edited Nov 16 at 9:27
MathFun123
516216
asked Nov 16 at 8:09
SuperMage1
852210
edited Nov 16 at 9:27
MathFun123
516216
asked Nov 16 at 8:09
SuperMage1
852210
edited Nov 16 at 9:27
MathFun123
516216
edited Nov 16 at 9:27
MathFun123
516216
edited Nov 16 at 9:27
MathFun123
516216
516216
asked Nov 16 at 8:09
SuperMage1
852210
asked Nov 16 at 8:09
SuperMage1
852210
asked Nov 16 at 8:09
SuperMage1
852210
852210
add a comment |
add a comment |
1 Answer
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2
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Let the centre of the circle be at the origin, and let the point have coordinates $(p,q)$. Then $p^2+(q+2)^2=50 text{(eq. 1)}$, and $(p-6)^2+q^2=50 text{(eq. 2)}$, so $p^2+(q+2)^2=(p-6)^2+q^2 text{(eq. 3)} $.
Expanding, we have:
$$p^2+q^2+4q+4=p^2-12p+36+q^2$$
$$4q+4=-12p+36$$
$$q+1=-3p+9$$
$$q=-3p+8 text{(eq. 4)}$$
You can then substitute back into equation $1$ to find $p$. Then you can use equation $2$ and subtract $-12p+36$ from the LHS to find $p^2+q^2$, then $sqrt{p^2+q^2}$.
answered Nov 16 at 8:46
Toby Mak
3,30311128
You might want to start off by saying, let the center of the circle be at the origin.
– Gerry Myerson
Nov 16 at 11:51
1
@GerryMyerson Fixed.
– Toby Mak
Nov 16 at 11:52
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let the centre of the circle be at the origin, and let the point have coordinates $(p,q)$. Then $p^2+(q+2)^2=50 text{(eq. 1)}$, and $(p-6)^2+q^2=50 text{(eq. 2)}$, so $p^2+(q+2)^2=(p-6)^2+q^2 text{(eq. 3)} $.
Expanding, we have:
$$p^2+q^2+4q+4=p^2-12p+36+q^2$$
$$4q+4=-12p+36$$
$$q+1=-3p+9$$
$$q=-3p+8 text{(eq. 4)}$$
You can then substitute back into equation $1$ to find $p$. Then you can use equation $2$ and subtract $-12p+36$ from the LHS to find $p^2+q^2$, then $sqrt{p^2+q^2}$.
answered Nov 16 at 8:46
Toby Mak
3,30311128
You might want to start off by saying, let the center of the circle be at the origin.
– Gerry Myerson
Nov 16 at 11:51
1
@GerryMyerson Fixed.
– Toby Mak
Nov 16 at 11:52
add a comment |
up vote
2
down vote
Let the centre of the circle be at the origin, and let the point have coordinates $(p,q)$. Then $p^2+(q+2)^2=50 text{(eq. 1)}$, and $(p-6)^2+q^2=50 text{(eq. 2)}$, so $p^2+(q+2)^2=(p-6)^2+q^2 text{(eq. 3)} $.
Expanding, we have:
$$p^2+q^2+4q+4=p^2-12p+36+q^2$$
$$4q+4=-12p+36$$
$$q+1=-3p+9$$
$$q=-3p+8 text{(eq. 4)}$$
You can then substitute back into equation $1$ to find $p$. Then you can use equation $2$ and subtract $-12p+36$ from the LHS to find $p^2+q^2$, then $sqrt{p^2+q^2}$.
answered Nov 16 at 8:46
Toby Mak
3,30311128
You might want to start off by saying, let the center of the circle be at the origin.
– Gerry Myerson
Nov 16 at 11:51
1
@GerryMyerson Fixed.
– Toby Mak
Nov 16 at 11:52
add a comment |
up vote
2
down vote
up vote
2
down vote
Let the centre of the circle be at the origin, and let the point have coordinates $(p,q)$. Then $p^2+(q+2)^2=50 text{(eq. 1)}$, and $(p-6)^2+q^2=50 text{(eq. 2)}$, so $p^2+(q+2)^2=(p-6)^2+q^2 text{(eq. 3)} $.
Expanding, we have:
$$p^2+q^2+4q+4=p^2-12p+36+q^2$$
$$4q+4=-12p+36$$
$$q+1=-3p+9$$
$$q=-3p+8 text{(eq. 4)}$$
You can then substitute back into equation $1$ to find $p$. Then you can use equation $2$ and subtract $-12p+36$ from the LHS to find $p^2+q^2$, then $sqrt{p^2+q^2}$.
answered Nov 16 at 8:46
Toby Mak
3,30311128
Let the centre of the circle be at the origin, and let the point have coordinates $(p,q)$. Then $p^2+(q+2)^2=50 text{(eq. 1)}$, and $(p-6)^2+q^2=50 text{(eq. 2)}$, so $p^2+(q+2)^2=(p-6)^2+q^2 text{(eq. 3)} $.
Expanding, we have:
$$p^2+q^2+4q+4=p^2-12p+36+q^2$$
$$4q+4=-12p+36$$
$$q+1=-3p+9$$
$$q=-3p+8 text{(eq. 4)}$$
You can then substitute back into equation $1$ to find $p$. Then you can use equation $2$ and subtract $-12p+36$ from the LHS to find $p^2+q^2$, then $sqrt{p^2+q^2}$.
answered Nov 16 at 8:46
Toby Mak
3,30311128
edited Nov 16 at 11:52
answered Nov 16 at 8:46
Toby Mak
3,30311128
answered Nov 16 at 8:46
Toby Mak
3,30311128
answered Nov 16 at 8:46
Toby Mak
3,30311128
3,30311128
You might want to start off by saying, let the center of the circle be at the origin.
– Gerry Myerson
Nov 16 at 11:51
1
@GerryMyerson Fixed.
– Toby Mak
Nov 16 at 11:52
add a comment |
You might want to start off by saying, let the center of the circle be at the origin.
– Gerry Myerson
Nov 16 at 11:51
1
@GerryMyerson Fixed.
– Toby Mak
Nov 16 at 11:52
You might want to start off by saying, let the center of the circle be at the origin.
– Gerry Myerson
Nov 16 at 11:51
You might want to start off by saying, let the center of the circle be at the origin.
– Gerry Myerson
Nov 16 at 11:51
1
1
@GerryMyerson Fixed.
– Toby Mak
Nov 16 at 11:52
@GerryMyerson Fixed.
– Toby Mak
Nov 16 at 11:52
add a comment |
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