The radius of a circle C_1 is thrice the radi | vedclass

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(D) Let the radius of $C_2$ be $r_2 = r$ and the radius of $C_1$ be $r_1 = 3r$. The centres are $O_1(1, 2)$ and $O_2(3, -2)$. The distance between the

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$x^2+y^2-2x+4y+3=0$

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(D) Let the radius of $C_2$ be $r_2 = r$ and the radius of $C_1$ be $r_1 = 3r$. The centres are $O_1(1, 2)$ and $O_2(3, -2)$. The distance between the centres $d$ is given by $d^2 = (3-1)^2 + (-2-2)^2 = 2^2 + (-4)^2 = 4 + 16 = 20$. Since the circles cut each other orthogonally,the condition is $d^2 = r_1^2 + r_2^2$. Substituting the values: $20 = (3r)^2 + r^2 = 9r^2 + r^2 = 10r^2$. Thus,$10r^2 = 20$,which implies $r^2 = 2$. The equation of a circle with centre $(h, k) = (1, -2)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values: $(x-1)^2 + (y+2)^2 = 2$. Expanding this: $x^2 - 2x + 1 + y^2 + 4y + 4 = 2$. $x^2 + y^2 - 2x + 4y + 5 = 2$. $x^2 + y^2 - 2x + 4y + 3 = 0$.

The radius of a circle $C_1$ is thrice the radius of another circle $C_2$. The centres of $C_1$ and $C_2$ are $(1, 2)$ and $(3, -2)$ respectively. If they cut each other orthogonally,find the equation of the circle with radius $r$ and centre $(1, -2)$.

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