Find the equation of the tangents to the circle $x^{2} + y^{2} - 22 x - 4 y + 25 = 0$ , which are perpendicular to the straight line $5 x + 12 y + 9 = 0$ .

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Solution

Equation of circle is $x^{2} + y^{2} - 22 x - 4 y + 25 = 0$
equation of st line is $5 x + 12 y + 9 = 0$
equation of line $⊥$ to it is $12 x - 5 y + λ = 0$ its from centre of given circle $=$ radius of the circle.
Centre of circle is $(- 11, - 2)$
radius $= \sqrt{121 + 16 - 25} = \sqrt{137 - 25}$
$= \sqrt{112} = \sqrt{4 \times 28}$
$= \sqrt{4 \times 4 \times 7} = 4 \sqrt{7}$
$\Rightarrow \frac{| 12 x - 11 + 12 x - 2 + 9 |}{\sqrt{25 + 144}} = 4 \sqrt{7}$
$\frac{| 12 x - 11 - 5 x - 2 + λ |}{\sqrt{25 + 144}} = 4 \sqrt{7}$
$\Rightarrow \frac{| - 132 + 10 + λ |}{13} = 4 \sqrt{7}$
$\frac{| λ - 122 |}{13} = 4 \sqrt{7} \Rightarrow λ - 122 = 52 \sqrt{5}$
$λ = 122 + 52 \sqrt{5}$
$\Rightarrow$ Equation of tangent is $12 x - 5 y + 122 + 52 \sqrt{5} = 0$

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