Find the equation of a circle with radius 4 t | vedclass

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(B) The circle touches the $x$-axis at the point $(-3, 0)$.Since the radius of the circle is $r = 4$,the center of the circle must be at $(-3, 4)$ or

Vedclass · Accepted Answer

$x^2 + y^2 + 6x \pm 8y + 9 = 0$

Vedclass · Answer

(B) The circle touches the $x$-axis at the point $(-3, 0)$.Since the radius of the circle is $r = 4$,the center of the circle must be at $(-3, 4)$ or $(-3, -4)$.Case $1$: Center is $(-3, 4)$ and radius $r = 4$.The equation is $(x - (-3))^2 + (y - 4)^2 = 4^2$.$(x + 3)^2 + (y - 4)^2 = 16$.$x^2 + 6x + 9 + y^2 - 8y + 16 = 16$.$x^2 + y^2 + 6x - 8y + 9 = 0$.Case $2$: Center is $(-3, -4)$ and radius $r = 4$.The equation is $(x - (-3))^2 + (y - (-4))^2 = 4^2$.$(x + 3)^2 + (y + 4)^2 = 16$.$x^2 + 6x + 9 + y^2 + 8y + 16 = 16$.$x^2 + y^2 + 6x + 8y + 9 = 0$.Combining both cases,the equation is $x^2 + y^2 + 6x \pm 8y + 9 = 0$.

Find the equation of a circle with radius $4$ that touches the $x$-axis at a distance of $-3$ from the origin.

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