If a circle passes through the points (2, 3) | vedclass

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(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

Vedclass · Accepted Answer

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

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 $(2, 3)$ and $(4, 5)$ is 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 $(2, 5)$.Radius squared $r^2 = (2-2)^2 + (5-3)^2 = 0 + 4 = 4$.The equation is $(x-2)^2 + (y-5)^2 = 4 \implies x^2 - 4x + 4 + y^2 - 10y + 25 = 4 \implies x^2 + y^2 - 4x - 10y + 25 = 0$.

If a circle passes through the points $(2, 3)$ and $(4, 5)$ and its center lies on the straight line $y - 4x + 3 = 0$, then its equation is:

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