If a circle passing through the point (1,1) c | vedclass

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(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+C=0 \quad (i)$.Since the circle passes through $(1,1)$,we have $1+1+2g+2f+C=0$,which implies $2

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$\left(\frac{3}{4}, \frac{5}{4}\right)$

Vedclass · Answer

(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+C=0 \quad (i)$.Since the circle passes through $(1,1)$,we have $1+1+2g+2f+C=0$,which implies $2g+2f+C=-2 \quad (ii)$.The condition for two circles $x^2+y^2+2g_1x+2f_1y+C_1=0$ and $x^2+y^2+2g_2x+2f_2y+C_2=0$ to be orthogonal is $2g_1g_2+2f_1f_2=C_1+C_2$.For the circle $x^2+y^2+4x-5=0$,$g_1=2, f_1=0, C_1=-5$. Orthogonality gives $2g(2)+2f(0)=C-5$,so $4g=C-5 \quad (iii)$.For the circle $x^2+y^2-4y+3=0$,$g_2=0, f_2=-2, C_2=3$. Orthogonality gives $2g(0)+2f(-2)=C+3$,so $-4f=C+3 \quad (iv)$.From $(iii)$,$C=4g+5$. From $(iv)$,$C=-4f-3$. Equating them: $4g+5=-4f-3$ $\Rightarrow 4g+4f=-8$ $\Rightarrow g+f=-2 \quad (v)$.Substituting $C=4g+5$ into $(ii)$: $2g+2f+4g+5=-2 \Rightarrow 6g+2f=-7$. Since $f=-2-g$,$6g+2(-2-g)=-7$ $\Rightarrow 4g=-3$ $\Rightarrow g=-\frac{3}{4}$.Then $f=-2-(-\frac{3}{4}) = -\frac{5}{4}$.The center of the circle is $(-g, -f) = \left(\frac{3}{4}, \frac{5}{4}\right)$.

If a circle passing through the point $(1,1)$ cuts the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-4y+3=0$ orthogonally,then the center of that circle is:

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