If (frac{1}{10}, frac{-1}{5}) is the inverse | vedclass

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(B) The inverse point $P'(x', y')$ of a point $P(x_1, y_1)$ with respect to a circle $S: x^2 + y^2 + 2gx + 2fy + c = 0$ lies on the line joining the c

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$-4$

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(B) The inverse point $P'(x', y')$ of a point $P(x_1, y_1)$ with respect to a circle $S: x^2 + y^2 + 2gx + 2fy + c = 0$ lies on the line joining the center $O(-g, -f)$ and the point $P$. Given the circle $x^2 + y^2 - 2x + 4y + c = 0$,the center is $O(1, -2)$. The point $P$ is $(-1, 2)$. The inverse point $P'$ is $(\frac{1}{10}, \frac{-1}{5})$. The power of point $P$ with respect to the circle is $S_1 = x_1^2 + y_1^2 - 2x_1 + 4y_1 + c = (-1)^2 + (2)^2 - 2(-1) + 4(2) + c = 1 + 4 + 2 + 8 + c = 15 + c$. For an inverse point,the relation is $OP \cdot OP' = r^2$,where $r^2 = g^2 + f^2 - c = 1^2 + (-2)^2 - c = 5 - c$. Using the property of inverse points,the point $P'$ must satisfy the condition that the power of the point $P$ is related to the circle. Alternatively,since $P'$ is the inverse of $P$,$P'$ lies on the line $OP$ and $OP \cdot OP' = r^2$. Vector $OP = (-1-1, 2-(-2)) = (-2, 4)$. Vector $OP' = (\frac{1}{10}-1, \frac{-1}{5}-(-2)) = (-\frac{9}{10}, \frac{9}{5})$. $OP \cdot OP' = (-2)(-\frac{9}{10}) + (4)(\frac{9}{5}) = \frac{18}{10} + \frac{36}{5} = 1.8 + 7.2 = 9$. Thus,$r^2 = 9$. Since $r^2 = 5 - c$,we have $5 - c = 9$,which gives $c = -4$.

If $(\frac{1}{10}, \frac{-1}{5})$ is the inverse point of a point $(-1, 2)$ with respect to the circle $x^2 + y^2 - 2x + 4y + c = 0$,then $c =$

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