The equation of the circle whose centre is (3 | vedclass

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Question

(A) Let $AB$ be the chord intercepted by the line from the circle and let $CD$ be the perpendicular drawn from the centre $C(3, -1)$ to the chord $AB$

Vedclass · Accepted Answer

$(x - 3)^2 + (y + 1)^2 = 38$

Vedclass · Answer

(A) Let $AB$ be the chord intercepted by the line from the circle and let $CD$ be the perpendicular drawn from the centre $C(3, -1)$ to the chord $AB$.Since the perpendicular from the centre bisects the chord,$AD = \frac{1}{2} AB = \frac{6}{2} = 3$.The length of the perpendicular $CD$ from the point $(3, -1)$ to the line $2x - 5y + 18 = 0$ is given by:$CD = \frac{|2(3) - 5(-1) + 18|}{\sqrt{2^2 + (-5)^2}} = \frac{|6 + 5 + 18|}{\sqrt{4 + 25}} = \frac{29}{\sqrt{29}} = \sqrt{29}$.In the right-angled triangle $	riangle CAD$,by Pythagoras theorem,the radius $r = CA$ is:$r^2 = CA^2 = AD^2 + CD^2 = 3^2 + (\sqrt{29})^2 = 9 + 29 = 38$.The equation of the circle with centre $(h, k) = (3, -1)$ and radius squared $r^2 = 38$ is:$(x - 3)^2 + (y - (-1))^2 = 38$,which simplifies to $(x - 3)^2 + (y + 1)^2 = 38$.

The equation of the circle whose centre is $(3, -1)$ and which cuts off a chord of length $6$ on the line $2x - 5y + 18 = 0$ is

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