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
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
The general equation of the circle is $\displaystyle x^2 + y^2 + 2gx + 2fy +K = 0$
The center of this circle is (-g, -f). If r is the radius of this circle, then
$\displaystyle (x+g)^2 + (y+f)^2 = r^2$
$\displaystyle x^2 + 2gx + g^2 + y^2 + 2fy + f^2 = r^2$
$\displaystyle x^2 + y^2 + 2gx + 2fy + g^2 + f^2 - r^2 = 0$
$\displaystyle x^2 + y^2 + 2gx + 2fy + k = 0$
For the second part of the question, it is yes.
Hello, Lukybear!
Your variables are confusing.
. . $\displaystyle d$ is a coordinate . . . and a distance ?
. . $\displaystyle x$ is a coordinate . . . and a variable ?
I must revise the problem.
$\displaystyle P(a,b)\text{ and }R(c,d) $ are points on: .$\displaystyle x^2 + y^2 + 2px + 2qy +r \:=\: 0$
$\displaystyle i)$ If $\displaystyle D$ is the distance between points $\displaystyle P$ and $\displaystyle R$, show that:
. . $\displaystyle -\frac{D^2}{2} \;=\;ac + bd + p(a+c) + q(b+d) + r$
$\displaystyle \begin{array}{ccccccc}P(a,b)\text{ is on the circle:} &a^2+b^2+2pa + 2qb + r &=& 0 \\ \\[-3mm]
R(c,d)\text{ is on the circle:} & c^2+d^2+2pc + 2qd + r &=& 0 \end{array}$
Add the equations: .$\displaystyle a^2+c^2+b^2+d^2+2pa + 2pc + 2qb + 2qd + 2r \:=\:0$
And we have: .$\displaystyle (a^2+c^2)+(b^2+d^2) + 2p(a+c) + 2q(b+d) + 2r \:=\:0$
. . Hence: .$\displaystyle (a^2+b^2) + (b^2+d^2) \;=\;-2\bigg[p(a+c) + q(b+d) + r\bigg]$ .[1]
The distance between $\displaystyle P$ and $\displaystyle R$ is given by: .$\displaystyle D^2 \;=\;(a-c)^2 + (b-d)^2)$
And we have: .$\displaystyle D^2 \;=\;a^2-2ac + c^2 + b^2 - 2bd + d^2$
. . Hence: .$\displaystyle D^2 \;=\;(a^2+c^2) + (b^2 + d^2) - 2(ac + bd)$ .[2]
Substitute [1] into [2]: .$\displaystyle D^2 \;=\;-2\bigg[p(a+c) + q(b+d) + r\bigg] - 2(ac+bd)$
And we have: .$\displaystyle D^2 \;=\;-2(ac+bd) - 2p(a+c) - 2q(b+d) - 2r$
. . Therefore: .$\displaystyle -\frac{D^2}{2} \;=\;ac+bd + p(a+c) + q(b+d) + r$