The equation of the circle whose centre lies | vedclass

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(D) Let the centre of the circle be $(h, k)$. Since it lies on the line $x-4y=1$,we have $h = 1+4k$. Thus,the centre is $(4k+1, k)$.Since the circle p

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$x^2+y^2+6x+2y-90=0$

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(D) Let the centre of the circle be $(h, k)$. Since it lies on the line $x-4y=1$,we have $h = 1+4k$. Thus,the centre is $(4k+1, k)$.Since the circle passes through $(3,7)$ and $(5,5)$,the distances from the centre to these points are equal (equal to the radius $r$):$(4k+1-3)^2 + (k-7)^2 = (4k+1-5)^2 + (k-5)^2$$(4k-2)^2 + (k-7)^2 = (4k-4)^2 + (k-5)^2$$16k^2 - 16k + 4 + k^2 - 14k + 49 = 16k^2 - 32k + 16 + k^2 - 10k + 25$$17k^2 - 30k + 53 = 17k^2 - 42k + 41$$12k = -12 \Rightarrow k = -1$.Substituting $k = -1$ into $h = 1+4k$,we get $h = 1+4(-1) = -3$. So,the centre is $(-3, -1)$.The radius $r$ is the distance between $(-3, -1)$ and $(5, 5)$:$r^2 = (5 - (-3))^2 + (5 - (-1))^2 = 8^2 + 6^2 = 64 + 36 = 100$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$:$(x + 3)^2 + (y + 1)^2 = 100$$x^2 + 6x + 9 + y^2 + 2y + 1 = 100$$x^2 + y^2 + 6x + 2y - 90 = 0$.

The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is

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