A circle C passing through the point (1, 1) b | vedclass

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(B) Let the equation of circle $C$ be $x^2+y^2+2gx+2fy+c=0$. Since $C$ passes through $(1, 1)$,we have $1+1+2g+2f+c=0$,which implies $2g+2f+c = -2$ (E

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$\left(\frac{5}{2}, 0\right)$

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(B) Let the equation of circle $C$ be $x^2+y^2+2gx+2fy+c=0$. Since $C$ passes through $(1, 1)$,we have $1+1+2g+2f+c=0$,which implies $2g+2f+c = -2$ (Equation $1$). Circle $C$ bisects the circumference of $x^2+y^2-2x=0$. The common chord is the radical axis $2gx+2fy+c - (-2x) = 0$,i.e.,$2(g+1)x+2fy+c=0$. For this to be a diameter of $x^2+y^2-2x=0$,it must pass through the centre $(1, 0)$. Substituting $(1, 0)$ into the radical axis equation: $2(g+1)(1)+2f(0)+c=0$,so $2g+c = -2$ (Equation $2$). Subtracting Equation $2$ from Equation $1$,we get $2f=0$,so $f=0$. Since $C$ is orthogonal to $x^2+y^2+2y-3=0$,we use the condition $2g_1g_2+2f_1f_2 = c_1+c_2$. Here $g_1=g, f_1=0, c_1=c$ and $g_2=0, f_2=1, c_2=-3$. So,$2(g)(0)+2(0)(1) = c-3$,which gives $c=3$. Substituting $c=3$ into Equation $2$: $2g+3 = -2$,so $2g = -5$,$g = -\frac{5}{2}$. The centre of circle $C$ is $(-g, -f) = (\frac{5}{2}, 0)$.

$A$ circle $C$ passing through the point $(1, 1)$ bisects the circumference of the circle $x^2+y^2-2x=0$. If $C$ is orthogonal to the circle $x^2+y^2+2y-3=0$,then the centre of the circle $C$ is

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