If the chord joining the points (1,2) and (2, | vedclass

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(A) Let the points be $A(1,2)$ and $B(2,-1)$. The length of the chord $AB$ is $L = \sqrt{(2-1)^2 + (-1-2)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{10}$.Let $R

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$x^2+y^2-6x-2y+5=0$

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(A) Let the points be $A(1,2)$ and $B(2,-1)$. The length of the chord $AB$ is $L = \sqrt{(2-1)^2 + (-1-2)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{10}$.Let $R$ be the radius of the circle. The angle subtended by the chord at the circumference is $	heta = \frac{\pi}{4}$.Using the formula $L = 2R \sin(	heta)$,we have $\sqrt{10} = 2R \sin(\frac{\pi}{4}) = 2R \cdot \frac{1}{\sqrt{2}} = R\sqrt{2}$.Thus,$R^2 = \frac{10}{2} = 5$.The equation of the circle is $(x-h)^2 + (y-k)^2 = 5$. Since $A$ and $B$ lie on the circle:$(1-h)^2 + (2-k)^2 = 5$ and $(2-h)^2 + (-1-k)^2 = 5$.Subtracting these equations: $(1-2h+h^2 + 4-4k+k^2) - (4-4h+h^2 + 1+2k+k^2) = 0 \implies 2h - 6k = 0 \implies h = 3k$.Substituting $h=3k$ into $(1-h)^2 + (2-k)^2 = 5$: $(1-3k)^2 + (2-k)^2 = 5 \implies 1-6k+9k^2 + 4-4k+k^2 = 5 \implies 10k^2 - 10k = 0$.So $k=0$ or $k=1$. If $k=0$,$h=0$ (not possible as $R^2=5$). If $k=1$,$h=3$. The center is $(3,1)$.The equation is $(x-3)^2 + (y-1)^2 = 5 \implies x^2-6x+9 + y^2-2y+1 = 5 \implies x^2+y^2-6x-2y+5=0$.

If the chord joining the points $(1,2)$ and $(2,-1)$ on a circle subtends an angle of $\frac{\pi}{4}$ at any point on its circumference,then the equation of such a circle is:

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