The equation of the circle with (1,1) as cent | vedclass

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(B) The centre of the circle is $(h, k) = (1, 1)$.The equation of the circle is $(x-1)^2 + (y-1)^2 = r^2$,where $r$ is the radius.The perpendicular di

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$x^2+y^2-2x-2y-10=0$

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(B) The centre of the circle is $(h, k) = (1, 1)$.The equation of the circle is $(x-1)^2 + (y-1)^2 = r^2$,where $r$ is the radius.The perpendicular distance $d$ from the centre $(1, 1)$ to the line $x+y+1=0$ is given by $d = \frac{|1+1+1|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$.The length of the chord is $L = 4\sqrt{2}$,so half the chord length is $a = \frac{L}{2} = 2\sqrt{2}$.Using the relation $r^2 = d^2 + a^2$,we get $r^2 = (\frac{3}{\sqrt{2}})^2 + (2\sqrt{2})^2 = \frac{9}{2} + 8 = \frac{9+16}{2} = \frac{25}{2} = 12.5$.The equation of the circle is $(x-1)^2 + (y-1)^2 = 12.5$,which simplifies to $x^2-2x+1 + y^2-2y+1 = 12.5$,or $x^2+y^2-2x-2y-10.5=0$.Multiplying by $2$,we get $2x^2+2y^2-4x-4y-21=0$.

The equation of the circle with $(1,1)$ as centre and which cuts a chord of length $4 \sqrt{2}$ units on the line $x+y+1=0$ is

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