The equation of a tangent to the circle $x^{2} + y^{2} = 25$ passing through $(- 2, 11)$ is

4 x + 3 y = 25

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3 x + 4 y = 38

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24 x - 7 y + 125 = 0

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7 x + 24 y = 230

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Solution

The correct options are
A $4 x + 3 y = 25$
C $24 x - 7 y + 125 = 0$
Circle = $x^{2} + y^{2} = 25$
Equation of tangent in slope form-
$y = m x \pm 5 \sqrt{1 + m^{2}}$
It passes throught (-2,11)
$11 = - 2 m \pm 5 \sqrt{1 + m^{2}}$
$(11 + 2 m) = \pm 5 \sqrt{1 + m^{2}}$
$(11 + 2 m)^{2} = 25 (1 + m^{2})$
$4 m^{2} + 44 m + 121 = 25 + 25 m^{2}$
$21 m^{2} - 44 m - 96 = 0$
$m = \frac{- 4}{3}, \frac{24}{7}$
Then tangent will be
$3 y + 4 x = 25$
Another tangent will be .
$7 y - 24 x = 125$
$24 x - 7 y + 125 = 0$
A,C are correct

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