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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Given the circle $x^2+y^2-4x+6y-12=0$, find the tangent at the point $(5,-7)$ on the circle.

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  • A tangent is perpendicular to the radius at the point of tangency.
  • The slope of the tangent $= \large\frac{-1}{\text{slope of radius}}$
  • Therefore the equation of the tangent can be obtained at the point of contact $(x_1,y_1)$ as $y-y_1 = m(x-x_1)$
Given a circle $x^2+y^2-4x+6y-12=0$, the centre of the circle = $(2,-3)$.
$\Rightarrow$ Slope of radius $ m_r = \large\frac{-7+3}{5-2} $$=\large\frac{-4}{3}$
$\Rightarrow$ Slope of tangent $m = \large\frac{-1}{m_r}$$ = \large\frac{-1\times 3}{-4}$$ = \large\frac{3}{4}$
Therefore the equation of the tangent at $(5,-7) = y-(-7) = \large\frac{3}{4} (x-5)$
$\qquad 4y + 28 = 3x - 15$
$\qquad 3x-4y-43=0$
answered Apr 9, 2014 by balaji.thirumalai
 

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