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  • The equation of the circle having centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is ______.

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

Answers (1)

The perpendicular distance from centre (3,-4) to the given line is,r=\left | \frac{5(3)+12(-4)-12}{\sqrt{25-144}} \right |

=\frac{45}{13}  which is radius of the circle

Required equation of the circle is \left ( x-3 \right )^{2}+\left ( y+4 \right )^{2}=\left (\frac{45}{13} \right )^{2}

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