Assertion: Consider a family of circles passing through two fixed points $A (3, 7)$ & $B (6, 5)$ and if the
chords in which the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ cuts the members of the family are
concurrent. Then the point of concurrency is $(2, \frac{23}{3})$
Reason: The equation of family of circles passing through A & B is
( $\bar{A B}$ diameter circle ) + $λ$ ( $\overset{\leftrightarrow}{A B}$ line) = 0

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(A) is false and (R) is true.

(A) is true and (R) is false.

Both (A) and (R) are true but (R) is not the correct explanation of (A).

Both (A) and (R) are true, and (R) is the correct explanation of (A).

answer is C.

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Detailed Solution

$\bar{A B}$ diameter circle is $(x - 3 (x - 6) + (y - 7) (y - 5) = 0$
$\Rightarrow x^{2} + y^{2} - 9 x - 12 y + 53 = 0$ and $\bar{A B}$ $2 x + 3 y = 27$
$∴$ The family of circles passing through A & B is $(x^{2} + y^{2} - 9 x - 12 y + 53) + λ (2 x + 3 y - 27) = 0$
But is intersected by $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ then the common chord is $(2 λ - 5) x + (3 λ - 6) y + 56 - 27 λ = 0$
By solving , we get $(2, \frac{23}{3})$