The circle with centre (2, 3) and intersecting $x^{2} + y^{2} - 4 x + 2 y - 7 = 0$ orthogonally has the radius

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 $2$
as Given $a n g l e C_{1} A C_{2} = 90^{0}$
$∴ r_{1}^{2} + r_{2}^{2} = (C_{1} C_{2})^{2}$
$r_{1}^{2} + 12 = 16$
$r_{1}^{2} = 4$
$r_{1} = 2$
$S : x^{2} + y^{2} - 4 x + 2 y - 7 = 0$

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