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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Given the circle $x^2+y^2+2x-4y+1=0$, find the length of a tangent from the point $(3,0)$ to the circle.

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Toolbox:
  • A tangent is perpendicular to the radius at the point of tangency.
  • Once we find the radius and distance from centre to the point we can use Pythogarous theorem to find the length of the tangent.
  • The radius of the circle is $\sqrt{g^2+f^2-c}$
Given a circle $x^2+y^2+2x-4y+1=0 \rightarrow$ centre of the circle $C=(-1,2)$
Given a point $A = (3,0)$, we need to find the length of $AB$ as depicted below:
The radius of the circle is $\sqrt{g^2+f^2-c}$
Radius $r = \sqrt{-1^2+2^2-1} = \sqrt 4 = 2$
Distance from centre of circle to point $A = \sqrt{(3+1)^2+(0-2)^2} = \sqrt{16+4} = \sqrt{20}$
$\Rightarrow$ Using Pythogarous Theorem, we can calculate $AB = \sqrt{AC^2 - BC^2} $
$\Rightarrow AB = \sqrt{\sqrt{20}^2 - 2^2} = \sqrt{20-4} = \sqrt{16} = 4$
answered Apr 9, 2014 by balaji.thirumalai
edited Apr 9, 2014 by balaji.thirumalai
 

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