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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Which of the following are tangents to the circle $x^2+y^2-6x-2y-15=0$ which are parallel to the line $3x+4y+20=0$?

$(1)\; 3x+4y-38 = 0$

$(2)\; 3x-4y-38 = 0$

$(3)\; 3x+4y+12 = 0$

$(4)\; 3x-4y-12 = 0$

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • The general equation of the circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$, where the centre is $(-g, -f)$
  • The radius of the circle is $\sqrt{g^2+f^2-c}$
  • The perpendicular distance from centre of circle to the tangent is equal to the radius.
Given the circle $x^2+y^2-6x-2y-15=0 \rightarrow$ centre $C = (3,1)$
Radius = $\sqrt{3^2+1^2+15} = \sqrt{25} = 5$
We have to find a tangent parallel to $3x+4y+20 = 0$. This will be of the form $3x+4y+k=0$, where $k$ is any constant.
Now, $3x+4y+k=0$ is a tangent to the circle, which means it is perpendicular to radius of length $5$ from $(3,1)$
$\Rightarrow 5 = \large\frac{ \left | 3(3) + 4(1) + k \right |}{\sqrt{3^2+4^2}}$$ = \large\frac{\left | 13+k \right |}{\sqrt{9+16}}$$ = \large \frac{\left| 13+k \right|}{5}$
$\Rightarrow \left|13+k\right| = 25 \rightarrow k = 12$ or $k=-38$
Therefore, the tangents are $3x+4y+12 = 0$ or $3x+4y-38 = 0$
answered Apr 9, 2014 by balaji.thirumalai
 

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