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Question

(A) Since the circle touches the $x$-axis at $(3, 0)$,its center is $(3, k)$ and its radius is $|k|$.The equation of the circle is $(x - 3)^2 + (y - k

Vedclass · Accepted Answer

$x^2 + y^2 - 6x - 5y + 9 = 0$

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(A) Since the circle touches the $x$-axis at $(3, 0)$,its center is $(3, k)$ and its radius is $|k|$.The equation of the circle is $(x - 3)^2 + (y - k)^2 = k^2$.Since it passes through $(1, 4)$,we substitute these coordinates into the equation:$(1 - 3)^2 + (4 - k)^2 = k^2$$(-2)^2 + 16 - 8k + k^2 = k^2$$4 + 16 - 8k = 0$$20 = 8k$$k = \frac{20}{8} = \frac{5}{2}$.Substituting $k = \frac{5}{2}$ back into the equation:$(x - 3)^2 + (y - \frac{5}{2})^2 = (\frac{5}{2})^2$$x^2 - 6x + 9 + y^2 - 5y + \frac{25}{4} = \frac{25}{4}$$x^2 + y^2 - 6x - 5y + 9 = 0$.

The equation of the circle which touches the $x$-axis at $(3, 0)$ and passes through $(1, 4)$ is given by

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