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

(C) infinitely many values of K

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Circle

If the circles

x^{2} + y^{2} + 5 K x + 2 y + K = 0

and

2 (x^{2} + y^{2}) + 2 K x + 3 y - 1 = 0, (K \in R)

, intersect at the points

P

and

Q

, then the line

4 x + 5 y - K = 0

passes through

P

and

Q

for:

(A)exactly one value of K
(B)no value of K✓ Correct
(C)infinitely many values of K
(D)exactly two values of K

Answer: (B)

Solution

The common chord of two circles is

S_{1} - S_{2} = 0

. Subtracting the given circles gives

4 K x + \frac{1}{2} y + K + \frac{1}{2} = 0

, or

8 K x + y + 2 K + 1 = 0

. For this to be the line

4 x + 5 y - K = 0

, the coefficient ratios must agree:

\frac{8 K}{4} = \frac{1}{5} = \frac{2 K + 1}{- K}

. The first equality needs

K = \frac{1}{10}

, but the second needs

K = - \frac{5}{11}

. These cannot both hold, so no value of

K

makes the line pass through both intersection points.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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