The number of circles having radius 5 and pas | vedclass

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Question

(B) Let the center of the circle be $(h, k)$.Since the circle passes through $(-2, 0)$ and $(4, 0)$,the center must lie on the perpendicular bisector

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the center of the circle be $(h, k)$.Since the circle passes through $(-2, 0)$ and $(4, 0)$,the center must lie on the perpendicular bisector of the segment joining these points.The midpoint of $(-2, 0)$ and $(4, 0)$ is $(\frac{-2+4}{2}, \frac{0+0}{2}) = (1, 0)$.The perpendicular bisector is the vertical line $x = 1$,so $h = 1$.The distance from the center $(1, k)$ to $(4, 0)$ is the radius $r = 5$.Using the distance formula: $\sqrt{(4-1)^2 + (0-k)^2} = 5$.$3^2 + k^2 = 5^2 \implies 9 + k^2 = 25$.$k^2 = 16 \implies k = \pm 4$.Thus,there are two possible centers: $(1, 4)$ and $(1, -4)$.Therefore,there are $2$ such circles.

The number of circles having radius $5$ and passing through the points $(-2, 0)$ and $(4, 0)$ is

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