Originally Posted by

**xwrathbringerx** Is this how u do it? Mayb it's wrong because the textbook's answer is actually $\displaystyle x^2 + y^2 + x - 3y = 0 $ instead of $\displaystyle x^2 + y^2 + 4x - 3y = 0$

Let pt (3,4) be A and pt (-1,2) be B.

AP is proportional to BP.

So $\displaystyle AP^2$is proportional to $\displaystyle BP^2$.

Therefore, AP = k × BP

$\displaystyle (x-3)^2 + (x-4)^2 = k ( (x+1)^2 + (x-2)^2)$

Substitute in (0,0)

$\displaystyle (0-3)^2 + (0-4)^2 = k( (0+1)^2 + (0-2)^2)$

9 + 16 = k (1+4)

25 = 5k

Therefore, k = 5

Substitute k = 5 into the original equation.

$\displaystyle (x-3)^2 + (x-4)^2 = 5 ( (x+1)^2 + (x-2)^2)$

$\displaystyle x^2 - 6x + 9 + y^2 - 8y + 16 = 5(x^2 + 2x + 1 + y^2 - 4y + 4)$

$\displaystyle x^2 + y^2 - 6x - 8y + 25 = 5(x^2 + 2x + y^2 - 4y + 5)$

$\displaystyle = 5x^2 + 10x + 5y^2 - 20y + 25 $

Hence,$\displaystyle x^2 + y^2 + 4x - 3y = 0$