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(A) Let the equation of the required circle be $(x - h)^{2} + (y - k)^{2} = r^{2}$.Since the circle passes through points $(4, 1)$ and $(6, 5)$:$(4 -

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$x^{2} + y^{2} - 6x - 8y + 15 = 0$

Vedclass · Answer

(A) Let the equation of the required circle be $(x - h)^{2} + (y - k)^{2} = r^{2}$.Since the circle passes through points $(4, 1)$ and $(6, 5)$:$(4 - h)^{2} + (1 - k)^{2} = r^{2}$  $(1)$$(6 - h)^{2} + (5 - k)^{2} = r^{2}$  $(2)$Since the centre $(h, k)$ lies on the line $4x + y = 16$:$4h + k = 16$  $(3)$Equating $(1)$ and $(2)$:$(4 - h)^{2} + (1 - k)^{2} = (6 - h)^{2} + (5 - k)^{2}$$16 - 8h + h^{2} + 1 - 2k + k^{2} = 36 - 12h + h^{2} + 25 - 10k + k^{2}$$17 - 8h - 2k = 61 - 12h - 10k$$4h + 8k = 44 \Rightarrow h + 2k = 11$  $(4)$Solving $(3)$ and $(4)$:From $(3)$,$k = 16 - 4h$. Substituting into $(4)$:$h + 2(16 - 4h) = 11$ $\Rightarrow h + 32 - 8h = 11$ $\Rightarrow -7h = -21$ $\Rightarrow h = 3$.Then $k = 16 - 4(3) = 4$.Substituting $h=3, k=4$ into $(1)$:$(4 - 3)^{2} + (1 - 4)^{2} = r^{2}$ $\Rightarrow 1^{2} + (-3)^{2} = r^{2}$ $\Rightarrow 1 + 9 = 10 = r^{2}$.The equation is $(x - 3)^{2} + (y - 4)^{2} = 10$.$x^{2} - 6x + 9 + y^{2} - 8y + 16 = 10 \Rightarrow x^{2} + y^{2} - 6x - 8y + 15 = 0$.

Find the equation of the circle passing through the points $(4, 1)$ and $(6, 5)$ and whose centre is on the line $4x + y = 16$.

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