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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections

The equation of the circle having centre at $(3,-4)$ and touching the line $5x+12y-12=0$ is __________

[Hint : To determine radius find the perpendicular distance from the centre of the circle to the line.]

$\begin{array}{1 1}(x-3)^2+(y+4)^2=45^2\\(x-3)^2+(y+4)^2=\large\frac{\sqrt{45}}{13}\\(x-3)^2+(y+4)^2=\big(\large\frac{45}{13}\big)^2\\(x-3)^2+(y+4)^2=\sqrt{\large\frac{45}{13}}\end{array} $

1 Answer

  • General equation of a circle having centre $(h,k)$ is $(x-h)^2+(y-k)^2=r^2$ where $r$ is the radius of the circle.
  • The perpendicular distance from the centre of the circle to the line touching the circle is the radius.(ie) $r=\bigg|\large\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\bigg|$ where $x_1$ and $y_1$ is the coordinates of the centre.
Answer : $(x-3)^2+(y+4)^2=\big(\large\frac{45}{13}\big)^2$
Given the coordinates of the centre is $(3,-4)$ and the equation of the line touching the circle is $5x+12y-12=0$
Hence $r=\bigg|\large\frac{5(3)+12(-4)-12}{\sqrt{25+144}}\bigg|$
$\Rightarrow r= \bigg|\large\frac{15-48-12}{13}\bigg|$
$\Rightarrow r= \large\frac{45}{13}$
Hence the equation of the circle is $(x-3)^2+(y+4)^2=\big(\large\frac{45}{13}\big)^2$
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