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(C) The pair of lines is given by $x^2 - y^2 - 2x + 4y - 3 = 0$. We can rewrite this as $x^2 - 2x - (y^2 - 4y) = 3$. Completing the square: $(x - 1)^2

Vedclass · Accepted Answer

$(x - 1)^2 + (y - 2)^2 = 1$

Vedclass · Answer

(C) The pair of lines is given by $x^2 - y^2 - 2x + 4y - 3 = 0$. We can rewrite this as $x^2 - 2x - (y^2 - 4y) = 3$. Completing the square: $(x - 1)^2 - 1 - ((y - 2)^2 - 4) = 3$,which simplifies to $(x - 1)^2 - (y - 2)^2 = 0$. This factors as $(x - 1 - (y - 2))(x - 1 + (y - 2)) = 0$,giving the lines $x - y + 1 = 0$ and $x + y - 3 = 0$. The center of the circle $(h, k)$ is the intersection of these two diameters: $h - k = -1$ and $h + k = 3$. Adding the equations gives $2h = 2$,so $h = 1$. Substituting $h = 1$ into $h + k = 3$ gives $k = 2$. The center is $(1, 2)$. The circle passes through $(1, 1)$,so the radius squared is $r^2 = (1 - 1)^2 + (2 - 1)^2 = 1$. The equation of the circle is $(x - 1)^2 + (y - 2)^2 = 1$.

The equation of the circle passing through the point $(1, 1)$ and having two diameters along the pair of lines $x^2 - y^2 - 2x + 4y - 3 = 0$ is

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