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(A) The radius $r$ of the circle is the perpendicular distance from the center $(3, 4)$ to the line $5x + 12y - 11 = 0$. $r = \left| \frac{5(3) + 12(4

Vedclass · Accepted Answer

$x^2 + y^2 - 6x - 8y + 9 = 0$

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(A) The radius $r$ of the circle is the perpendicular distance from the center $(3, 4)$ to the line $5x + 12y - 11 = 0$. $r = \left| \frac{5(3) + 12(4) - 11}{\sqrt{5^2 + 12^2}} \right| = \left| \frac{15 + 48 - 11}{\sqrt{25 + 144}} \right| = \frac{52}{13} = 4$. The equation of the circle with center $(h, k) = (3, 4)$ and radius $r = 4$ is given by $(x - h)^2 + (y - k)^2 = r^2$. $(x - 3)^2 + (y - 4)^2 = 4^2$ $x^2 - 6x + 9 + y^2 - 8y + 16 = 16$ $x^2 + y^2 - 6x - 8y + 9 = 0$.

The equation of the circle which has its centre at the point $(3, 4)$ and touches the line $5x + 12y - 11 = 0$ is

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