Find the equation of the circle with center ( | vedclass

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(C) The equation of a circle with center $(3, 1)$ is $(x - 3)^2 + (y - 1)^2 = r^2$.Since the circle touches the line $8x - 15y + 25 = 0$,the radius $r

Vedclass · Accepted Answer

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

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(C) The equation of a circle with center $(3, 1)$ is $(x - 3)^2 + (y - 1)^2 = r^2$.Since the circle touches the line $8x - 15y + 25 = 0$,the radius $r$ is equal to the perpendicular distance from the center $(3, 1)$ to the line.$r = \frac{|8(3) - 15(1) + 25|}{\sqrt{8^2 + (-15)^2}}$$r = \frac{|24 - 15 + 25|}{\sqrt{64 + 225}}$$r = \frac{|34|}{\sqrt{289}} = \frac{34}{17} = 2$.Substituting $r = 2$ into the circle equation:$(x - 3)^2 + (y - 1)^2 = 2^2$$(x^2 - 6x + 9) + (y^2 - 2y + 1) = 4$$x^2 + y^2 - 6x - 2y + 10 = 4$$x^2 + y^2 - 6x - 2y + 6 = 0$.

Find the equation of the circle with center $(3, 1)$ and touching the line $8x - 15y + 25 = 0$.

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