A circle passes through the points (1,2) and | vedclass

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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The centre of the circle is $(-g, -f)$. Since the centre lies on the line $x-y+3=0$,we ha

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$2$

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(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The centre of the circle is $(-g, -f)$. Since the centre lies on the line $x-y+3=0$,we have $-g - (-f) + 3 = 0$,which implies $f-g+3=0$ or $g = f+3$. The circle passes through $(1,2)$,so $1^2+2^2+2g(1)+2f(2)+c=0$,which simplifies to $2g+4f+c = -5$ (Equation $1$). The circle passes through $(3,4)$,so $3^2+4^2+2g(3)+2f(4)+c=0$,which simplifies to $6g+8f+c = -25$ (Equation $2$). Subtracting Equation $1$ from Equation $2$: $(6g-2g) + (8f-4f) = -25 - (-5) \implies 4g+4f = -20 \implies g+f = -5$ (Equation $3$). Substituting $g = f+3$ into Equation $3$: $(f+3)+f = -5 \implies 2f = -8 \implies f = -4$. Then $g = -4+3 = -1$. Substituting $g$ and $f$ into Equation $1$: $2(-1)+4(-4)+c = -5 \implies -2-16+c = -5 \implies c = 13$. The radius $r$ is given by $\sqrt{g^2+f^2-c} = \sqrt{(-1)^2+(-4)^2-13} = \sqrt{1+16-13} = \sqrt{4} = 2$.

$A$ circle passes through the points $(1,2)$ and $(3,4)$. If its centre lies on the line $x-y+3=0$,then its radius is equal to

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