The circumference of the circle x2+y2-2 x+8 y-q=0 is bisected by the circle x2+y2+4 x+12 y+p=0, then p+q is equal to:

Question

The circumference of the circle x2+y2-2 x+8 y-q=0 is bisected by the circle x2+y2+4 x+12 y+p=0, then p+q is equal to: (A) 25 (B) 100 (C) 10 (D) 48

Tardigrade Admin · Accepted Answer

Common chord of given circle 6 x+4 y+(p+q)=0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/bfd0091e0c29f68d5d6e5d6b14b5bf6a-.png" /> This is diameter of x2+y2-2 x+8 y-q=0 centre (1,-4) 6-16+(p+q)=0 ⇒ p+q=10

Q. The circumference of the circle $x^{2} + y^{2} - 2 x + 8 y - q = 0$ is bisected by the circle $x^{2} + y^{2} + 4 x + 12 y + p = 0$ , then $p + q$ is equal to:

25

100

10

48

Common chord of given circle
$6 x + 4 y + (p + q) = 0$

This is diameter of $x^{2} + y^{2} - 2 x + 8 y - q = 0$ centre $(1, - 4)$
$6 - 16 + (p + q) = 0 \Rightarrow p + q = 10$

Q. The circumference of the circle x2+y2−2x+8y−q=0 is bisected by the circle x2+y2+4x+12y+p=0, then p+q is equal to:

25

100

10

48

Common chord of given circle 6x+4y+(p+q)=0 This is diameter of x2+y2−2x+8y−q=0 centre (1,−4) 6−16+(p+q)=0⇒p+q=10

Q. The circumference of the circle $x^{2} + y^{2} - 2 x + 8 y - q = 0$ is bisected by the circle $x^{2} + y^{2} + 4 x + 12 y + p = 0$ , then $p + q$ is equal to:

Common chord of given circle
$6 x + 4 y + (p + q) = 0$

This is diameter of $x^{2} + y^{2} - 2 x + 8 y - q = 0$ centre $(1, - 4)$
$6 - 16 + (p + q) = 0 \Rightarrow p + q = 10$