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Question

(A) Let the equation of the circle be $(x - h)^2 + (y - k)^2 = r^2$.Since the circle passes through $(2, -2)$ and $(3, 4)$,we have:$(2 - h)^2 + (-2 -

Vedclass · Accepted Answer

$(x - 0.7)^2 + (y - 1.3)^2 = 12.58$

Vedclass · Answer

(A) Let the equation of the circle be $(x - h)^2 + (y - k)^2 = r^2$.Since the circle passes through $(2, -2)$ and $(3, 4)$,we have:$(2 - h)^2 + (-2 - k)^2 = r^2$  --- $(1)$$(3 - h)^2 + (4 - k)^2 = r^2$  --- $(2)$Also,since the centre $(h, k)$ lies on the line $x + y = 2$,we have:$h + k = 2$  --- $(3)$Equating $(1)$ and $(2)$:$(2 - h)^2 + (-2 - k)^2 = (3 - h)^2 + (4 - k)^2$$4 - 4h + h^2 + 4 + 4k + k^2 = 9 - 6h + h^2 + 16 - 8k + k^2$$8 - 4h + 4k = 25 - 6h - 8k$$2h + 12k = 17$  --- $(4)$From $(3)$,$h = 2 - k$. Substituting into $(4)$:$2(2 - k) + 12k = 17$$4 - 2k + 12k = 17$$10k = 13 \implies k = 1.3$$h = 2 - 1.3 = 0.7$Now,find $r^2$ using $(1)$:$r^2 = (2 - 0.7)^2 + (-2 - 1.3)^2 = (1.3)^2 + (-3.3)^2 = 1.69 + 10.89 = 12.58$Thus,the equation of the circle is $(x - 0.7)^2 + (y - 1.3)^2 = 12.58$.

Find the equation of the circle which passes through the points $(2, -2)$ and $(3, 4)$ and whose centre lies on the line $x + y = 2$.

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