# Thread: Conics - Distance between Two Points on Circle

1. ## Conics - Distance between Two Points on Circle

R (x, y) P (a,b) Q (c,d) are points on x^2 + y^2 + 2gx + 2fy +k = 0

i) if d is the distance between points R and P, show that:
-d^2/2 = x.a + y.b + g(x+a) + f(y+b) + k

I am not sure how to begin this question. Any approach is possible. Thanks guys

2. Originally Posted by Lukybear
R (x, y) P (a,b) Q (c,d) are points on x^2 + y^2 + 2gx + 2fy +k = 0

i) if d is the distance between points R and P, show that:
-d^2/2 = x.a + y.b + g(x+a) + f(y+b) + k

I am not sure how to begin this question. Any approach is possible. Thanks guys
Mid point of RP is ( $\frac{x+a}{2}, \frac{y+b}{2})$

If ( -g, -f ) is the center of the circle,

then $(\frac{d}{2})^2$ = $[(a+g)^2 + (b+f)^2] - [(\frac{x+a}{2} + g)^2 + (\frac{y+b}{2} + f)^2]$

Simplify and proceed.

3. Could you just expand on how the centre of circle is acquired?

Also, can i just acquire that the method used was by pythagoras? i.e. mid pt to centre is perpendicular to PR?

Thanks.

4. Originally Posted by Lukybear
Could you just expand on how the centre of circle is acquired?

Also, can i just acquire that the method used was by pythagoras? i.e. mid pt to centre is perpendicular to PR?

Thanks.
The general equation of the circle is $x^2 + y^2 + 2gx + 2fy +K = 0$

The center of this circle is (-g, -f). If r is the radius of this circle, then

$(x+g)^2 + (y+f)^2 = r^2$

$x^2 + 2gx + g^2 + y^2 + 2fy + f^2 = r^2$

$x^2 + y^2 + 2gx + 2fy + g^2 + f^2 - r^2 = 0$

$x^2 + y^2 + 2gx + 2fy + k = 0$

For the second part of the question, it is yes.

5. Hello, Lukybear!

. . $d$ is a coordinate . . . and a distance ?
. . $x$ is a coordinate . . . and a variable ?

I must revise the problem.

$P(a,b)\text{ and }R(c,d)$ are points on: . $x^2 + y^2 + 2px + 2qy +r \:=\: 0$

$i)$ If $D$ is the distance between points $P$ and $R$, show that:

. . $-\frac{D^2}{2} \;=\;ac + bd + p(a+c) + q(b+d) + r$

$\begin{array}{ccccccc}P(a,b)\text{ is on the circle:} &a^2+b^2+2pa + 2qb + r &=& 0 \\ \\[-3mm]
R(c,d)\text{ is on the circle:} & c^2+d^2+2pc + 2qd + r &=& 0 \end{array}$

Add the equations: . $a^2+c^2+b^2+d^2+2pa + 2pc + 2qb + 2qd + 2r \:=\:0$

And we have: . $(a^2+c^2)+(b^2+d^2) + 2p(a+c) + 2q(b+d) + 2r \:=\:0$

. . Hence: . $(a^2+b^2) + (b^2+d^2) \;=\;-2\bigg[p(a+c) + q(b+d) + r\bigg]$ .[1]

The distance between $P$ and $R$ is given by: . $D^2 \;=\;(a-c)^2 + (b-d)^2)$

And we have: . $D^2 \;=\;a^2-2ac + c^2 + b^2 - 2bd + d^2$

. . Hence: . $D^2 \;=\;(a^2+c^2) + (b^2 + d^2) - 2(ac + bd)$ .[2]

Substitute [1] into [2]: . $D^2 \;=\;-2\bigg[p(a+c) + q(b+d) + r\bigg] - 2(ac+bd)$

And we have: . $D^2 \;=\;-2(ac+bd) - 2p(a+c) - 2q(b+d) - 2r$

. . Therefore: . $-\frac{D^2}{2} \;=\;ac+bd + p(a+c) + q(b+d) + r$

6. Wow thats brilliant. Thxs very much! Do apologise for question.