Question

For different values of k, the circle $x^2+y^2$ + (8 + k)x + (8 + k)y + (16 + 12k) = 0, always passes through two fixed point P and Q. For $k=k_1$, the tangents at P and Q intersect at the origin. Which of the following is/are correct?

• A. the mid-point of P and Q is (–6, –6)
• B. the sum of ordinates of P and Q is –12
• C. $k_1$ may be equal to $–\frac{32}{9}$
• D. $k_1$ may be equal to $\frac{8}{3}$

Answer 1

Answer: (A), (B)

The correct options are
A

the mid-point of P and Q is (–6, –6)

B

the sum of ordinates of P and Q is –12

Circle is $x^2+y^2+(8+k)x+(8+k)y+(16+12k)=0$

Or, $x^2+y^2+8x+8y+16+k(x+y+12)=0$

From origin chord of contact is T = 0

$\Rightarrow (8+k_1)x+(8+k_1)y+8(4+3k_1)=0$

Which is same as $x+y+12=0$

$\Rightarrow \frac{8+k_1}{1}=\frac{8(4+3k_1)}{12}\Rightarrow k_1=\frac{16}{3}$

As PQ is chord to $x^2+y^2+8x+8y+16+\frac{16}{3}(x+y+12)=0$

From mid-point of PQ(h, k) chord is T = $S_1$ $\Rightarrow$ (h, k) = (–6, –6)