A circle passes through the centre of another | vedclass

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(A) The centre of the circle $x^2+y^2-3x-4y-1=0$ is $C(\frac{3}{2}, 2)$.Since the required circle passes through the point $C(\frac{3}{2}, 2)$ and has

Vedclass · Accepted Answer

$4x^2+4y^2-40x-16y+67=0$

Vedclass · Answer

(A) The centre of the circle $x^2+y^2-3x-4y-1=0$ is $C(\frac{3}{2}, 2)$.Since the required circle passes through the point $C(\frac{3}{2}, 2)$ and has its centre at $(5, 2)$,the radius $r$ of the required circle is the distance between these two points:$r = \sqrt{(5 - \frac{3}{2})^2 + (2 - 2)^2} = \sqrt{(\frac{7}{2})^2} = \frac{7}{2}$.The equation of the required circle is $(x-5)^2 + (y-2)^2 = r^2$.$(x-5)^2 + (y-2)^2 = (\frac{7}{2})^2$$x^2 - 10x + 25 + y^2 - 4y + 4 = \frac{49}{4}$$x^2 + y^2 - 10x - 4y + 29 = \frac{49}{4}$Multiplying by $4$ on both sides:$4x^2 + 4y^2 - 40x - 16y + 116 = 49$$4x^2 + 4y^2 - 40x - 16y + 67 = 0$.Thus,option $A$ is correct.

$A$ circle passes through the centre of another circle $x^2+y^2-3x-4y-1=0$ and its centre is $(5,2)$. Then the equation of this circle is.........

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