The intercept made by the circle $x^{2} + y^{2} - 14 x - 10 y + 23 = 0$ on the line $7 x + y - 4 = 0$ is

1

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2

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3

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4

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Solution

The correct option is B $2$
center of circle $= (7, 5)$
radius $= \sqrt{7^{2} + 5^{2} - 23} = \sqrt{51}$
Distance of centre from given line is
$\frac{7 (7) + (5) - 4}{\sqrt{7^{2} + 12}} = \sqrt{50}$
Now, $A M = M B$
$A B = i n t e r c e p t$
$A M = \sqrt{(A O)^{2} - (M O)^{2}} = \sqrt{51 - 50}$
$A M = 1$
$∴ A B = A M + M B = 2 (A M)$
$= 2 =$ intercept

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