The area of the circle having its centre at ( | vedclass

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(A) Given,the centre of the circle is $C(3,4)$.Since the circle touches the line $5x+12y-11=0$,the radius $r$ of the circle is equal to the perpendicu

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$16 \pi$ sq units

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(A) Given,the centre of the circle is $C(3,4)$.Since the circle touches the line $5x+12y-11=0$,the radius $r$ of the circle is equal to the perpendicular distance from the centre $(3,4)$ to the line.The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $ax+by+c=0$ is given by $d = \frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$.Substituting the values,we get:$r = \frac{|5(3)+12(4)-11|}{\sqrt{5^2+12^2}}$$r = \frac{|15+48-11|}{\sqrt{25+144}}$$r = \frac{|52|}{\sqrt{169}}$$r = \frac{52}{13} = 4$.The area of the circle is given by $\pi r^2$.Area $= \pi(4)^2 = 16\pi$ sq units.

The area of the circle having its centre at $(3,4)$ and touching the line $5x+12y-11=0$ is

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