Statement 1 - If the circumference of the circle $x^{2} + y^{2} - 2 x + 8 y - q^{2} = 0$ is bisected by the circle $x^{2} + y^{2} + 4 x + 22 y + p^{2} = 0$ , then $pq$ cannot exceed 25.

Statement 2 - The common chord of two circles must be equidistant from the centres of both the circles.

Question

Statement 1 - If the circumference of the circle $x^{2} + y^{2} - 2 x + 8 y - q^{2} = 0$ is bisected by the circle $x^{2} + y^{2} + 4 x + 22 y + p^{2} = 0$ , then $pq$ cannot exceed 25.

Statement 2 - The common chord of two circles must be equidistant from the centres of both the circles.

TestHub · Accepted Answer

The common chord is 6 x + 14 y + p 2 + q 2 = 0 . It is a diameter of the first circle, so it passes through ( 1 , − 4 ) : 6 − 56 + p 2 + q 2 = 0 , p 2 + q 2 = 50 . Then pq ≤ 2 p 2 + q 2 ​ = 25 , so Statement 1 is true. The common chord is equidistant from the centres only when the circles have equal radii, so Statement 2 is false.

Mathematics - Circle Question with Solution | TestHub

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