Question

If P and Q are the points of intersection of the circles $x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ and $x^2 + y^2 + 2x + 2y - p^2 = 0$ , then there is a circle passing through $P, Q$ and $(1, 1)$ for

• A. all values of p
• B. all except one value of p
• C. all except two values of p
• D. exactly one value of p

Answer 1

Answer: (B)

all except one value of p

Answer 2

Given circles $S = x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ $S' = x^2 + y^2 + 2x + 2y - p^2 = 0$ Equation of required circle is $S+\lambda S '=0$ As it passes through $\left(1, 1\right)$ the value of $\lambda=\frac{-\left(7+2p\right)}{\left(6-p^{2}\right)}M$ If $7 + 2p = 0$, it becomes the second circle $\therefore$ it is true for all values of $p$