The centre of the circle which intersects the circle $x^2+y^2-2 x-2 y-2=0$ orthogonally and passes through the point $(2,0)$ and touches the $X$-axis is

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Aicharya · Accepted Answer

Let equation of circle,

Whose centre is $=(-g,-f)$ Circle passes through the point $(2,0)$, so this point satisfy the circle. $ 4+0+4 g+0+c=0 $

From Eqs. (ii) and (iii), we get $ \begin{array}{rlrl} 4+4 g+g^2 & =0 \\ (g+2)^2=0 \Rightarrow g & =-2 \\ \Rightarrow \quad & & c & =4 \end{array} $ Now, circle Eq. (i) intersect the circle $ x^2+y^2-2 x-2 y-2=0 \text { orthogonally } $ So, condition of orthogonality $ \begin{aligned} 2 g_1 g_2+2 f_1 f_2 & =c_1+c_2 \\ 2(-2)(-1)+2 f \cdot(-1) & =4-2 \\ 4-2 f & =2 \\ 4-2 & =2 f \\ \Rightarrow \quad f & =1 \end{aligned} $ Hence, coordinate of centre is $(2,-1)$

The centre of the circle which intersects the circle $x^2+y^2-2 x-2 y-2=0$ orthogonally and passes through…

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