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Equation of Normal at Given Point

Find the leng...

Question

Find the length of tangent segment of the circle $4 x^{2} + 4 y^{2} + 16 x - 24 y + 3 = 0$ from the point $(- 1, - 1)$ .

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Solution

Length of the tangent segment of the circle
$4 x^{2} + 4 y^{2} + 16 x - 24 y + 3 = 0$
$x^{2} + y^{2} + 4 x - 6 y + \frac{3}{4} = 0$ from the point $(- 1, - 1)$ .
$= \sqrt{{(- 1)}^{2} + {(- 1)}^{2} + 4 (- 1) - 6 (- 1) + \frac{3}{4}}$
$= \sqrt{1 + 1 - 4 + 6 + \frac{3}{4}}$
$= \sqrt{\frac{19}{4}} = \frac{\sqrt{19}}{2}$ units

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