A circle passing through the points (1, 1) an | vedclass

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(A) The points $(1, 1)$ and $(2, 0)$ lie on the circle $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $(1, 1)$: $1 + 1 + 2g + 2f + c = 0 \Rightarrow 2g

Vedclass · Accepted Answer

$-\frac{5}{2}$

Vedclass · Answer

(A) The points $(1, 1)$ and $(2, 0)$ lie on the circle $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $(1, 1)$: $1 + 1 + 2g + 2f + c = 0 \Rightarrow 2g + 2f + c = -2$  ...$(i)$Substituting $(2, 0)$: $4 + 0 + 4g + 0 + c = 0 \Rightarrow 4g + c = -4$  ...$(ii)$Subtracting $(i)$ from $(ii)$: $(4g + c) - (2g + 2f + c) = -4 - (-2)$ $\Rightarrow 2g - 2f = -2$ $\Rightarrow f = g + 1$.From $(ii)$,$c = -4 - 4g$.The center of the circle is $(-g, -f) = (-g, -(g + 1))$ and the radius $r$ is $\sqrt{g^2 + f^2 - c} = \sqrt{g^2 + (g + 1)^2 - (-4 - 4g)} = \sqrt{2g^2 + 6g + 5}$.The distance from the center $(-g, -g - 1)$ to the line $3x - y - 1 = 0$ is equal to the radius $r$:$\left|\frac{3(-g) - (-g - 1) - 1}{\sqrt{3^2 + (-1)^2}}\right| = r$ $\Rightarrow \left|\frac{-3g + g + 1 - 1}{\sqrt{10}}\right| = \sqrt{2g^2 + 6g + 5}$.$\left|\frac{-2g}{\sqrt{10}}\right| = \sqrt{2g^2 + 6g + 5} \Rightarrow \frac{4g^2}{10} = 2g^2 + 6g + 5$.$2g^2 = 10g^2 + 30g + 25 \Rightarrow 8g^2 + 30g + 25 = 0$.$(4g + 5)(2g + 5) = 0 \Rightarrow g = -\frac{5}{4}$ or $g = -\frac{5}{2}$.

$A$ circle passing through the points $(1, 1)$ and $(2, 0)$ touches the line $3x - y - 1 = 0$. If the equation of this circle is $x^2 + y^2 + 2gx + 2fy + c = 0$,then a possible value of $g$ is

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