Find the equation of a circle whose radius is | vedclass

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(A) The circle passes through points $(4, 0)$ and $(-4, 0)$ on the $x$-axis. Since the points are symmetric about the $y$-axis,the center of the circl

Vedclass · Accepted Answer

$x^2+y^2-6y-16=0$

Vedclass · Answer

(A) The circle passes through points $(4, 0)$ and $(-4, 0)$ on the $x$-axis. Since the points are symmetric about the $y$-axis,the center of the circle must lie on the $y$-axis. Let the center be $(0, c)$. The distance from the center $(0, c)$ to the point $(4, 0)$ is the radius $r = 5$. Using the distance formula: $\sqrt{(4-0)^2 + (0-c)^2} = 5$. $16 + c^2 = 25$ $\Rightarrow c^2 = 9$ $\Rightarrow c = \pm 3$. Thus,the centers are $(0, 3)$ and $(0, -3)$. For center $(0, 3)$,the equation is $x^2 + (y-3)^2 = 5^2$ $\Rightarrow x^2 + y^2 - 6y + 9 = 25$ $\Rightarrow x^2 + y^2 - 6y - 16 = 0$. For center $(0, -3)$,the equation is $x^2 + (y+3)^2 = 5^2$ $\Rightarrow x^2 + y^2 + 6y + 9 = 25$ $\Rightarrow x^2 + y^2 + 6y - 16 = 0$. Both $x^2 + y^2 - 6y - 16 = 0$ and $x^2 + y^2 + 6y - 16 = 0$ are valid equations.

Find the equation of a circle whose radius is $5$ units and passes through two points on the $x$-axis which are at a distance of $4$ units from the origin.

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