If the circles x^2+y^2+5Kx+2y+K=0 and 2(x^2+y^2)+2Kx+3y-1=0, (K R), intersect at the point…

Question

If the circles $x^2+y^2+5Kx+2y+K=0$ and $2(x^2+y^2)+2Kx+3y-1=0$, ($K\in\mathbb R$), intersect at the points $P$ and $Q$, then the line $4x+5y-K=0$ passes through $P$ and $Q$, for:

Accepted Answer

S1 $$ \equiv $$ x2 + y2 + 5Kx + 2y + K = 0

$${S_2} \equiv {x^2} + {y^2} + Kx + {3 \over 2}y - {1 \over 2} = 0$$

Equation of common chord is

S1 -- S2 = 0

$$ \Rightarrow 4Kx + {y \over 2} + K + {1 \over 2} = 0$$ ....(1)

4x + 5y -- K = 0 \ldots (2) (given)

On comparing (1) and (2)

$${{4K} \over 4} = {1 \over {10}} = {{2K + 1} \over { - 2K}}$$

$$ \Rightarrow K = {1 \over {10}}$$ and  $- 2K = 20K + 10$

$$ \Rightarrow $$ 22K = --10

$$\therefore K = {{ - 5} \over {11}}$$

So No value of K exists.

Solutions · 1