The equation of the circle which passes throu | vedclass

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Question

(B) Let the centre of the circle be $(h, k)$. Since the circle passes through $(2, 3)$ and $(4, 5)$,the distances from the centre to these points are

Vedclass · Accepted Answer

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

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(B) Let the centre of the circle be $(h, k)$. Since the circle passes through $(2, 3)$ and $(4, 5)$,the distances from the centre to these points are equal (equal to the radius $r$): $(h - 2)^2 + (k - 3)^2 = (h - 4)^2 + (k - 5)^2$ $h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 8h + 16 + k^2 - 10k + 25$ $4h + 4k = 28 \implies h + k = 7$ Given that the centre lies on $y - 4x + 3 = 0$,we have $k - 4h + 3 = 0$. Solving $h + k = 7$ and $k - 4h = -3$: Subtracting the equations: $(h + k) - (k - 4h) = 7 - (-3) \implies 5h = 10 \implies h = 2$. Then $k = 7 - 2 = 5$. The centre is $(2, 5)$. The radius squared is $r^2 = (2 - 2)^2 + (5 - 3)^2 = 0^2 + 2^2 = 4$. The equation of the circle is $(x - 2)^2 + (y - 5)^2 = 4$. $x^2 - 4x + 4 + y^2 - 10y + 25 = 4$ $x^2 + y^2 - 4x - 10y + 25 = 0$.

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

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