The equation of a circle which passes through | vedclass

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Question

(A) Let the center of the circle be $(h, k)$. Since the center lies on $y - 4x + 3 = 0$, we have $k - 4h + 3 = 0$, or $k = 4h - 3$.The distance from t

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$x^2 + y^2 - 4x - 10y + 25 = 0$

Vedclass · Answer

(A) Let the center of the circle be $(h, k)$. Since the center lies on $y - 4x + 3 = 0$, we have $k - 4h + 3 = 0$, or $k = 4h - 3$.The distance from the center $(h, k)$ to points $(2, 3)$ and $(4, 5)$ must be equal (radius $r$):$(h - 2)^2 + (k - 3)^2 = (h - 4)^2 + (k - 5)^2$$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 8h + 16 + k^2 - 10k + 25$$-4h - 6k + 13 = -8h - 10k + 41$$4h + 4k = 28 \implies h + k = 7$.Substitute $k = 4h - 3$ into $h + k = 7$:$h + (4h - 3) = 7 \implies 5h = 10 \implies h = 2$.Then $k = 4(2) - 3 = 5$.The center is $(2, 5)$.The radius squared is $r^2 = (2 - 2)^2 + (5 - 3)^2 = 0 + 4 = 4$.The equation is $(x - 2)^2 + (y - 5)^2 = 4$.$x^2 - 4x + 4 + y^2 - 10y + 25 = 4 \implies x^2 + y^2 - 4x - 10y + 25 = 0$.

The equation of a circle which passes through the points $(2, 3)$ and $(4, 5)$ and whose center lies on the straight line $y - 4x + 3 = 0$ is

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