Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Conic Sections
0 votes

Given the circle $x^2+y^2-4x+6y-12=0$, find the tangent at the point $(5,-7)$ on the circle.

Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App