If C(alpha, beta) with alpha

Question

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. The centre is $(-g, -f) = (\alpha, \beta)$.Since the circle touches the $Y$-axi

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$(-\sqrt{10}, 3)$

Vedclass · Answer

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. The centre is $(-g, -f) = (\alpha, \beta)$.Since the circle touches the $Y$-axis at $(0, 3)$,the radius is equal to the distance from the centre to the $Y$-axis,so $|\alpha| = |g| = r$. Also,$(0, 3)$ satisfies the equation: $0^2 + 3^2 + 2g(0) + 2f(3) + c = 0 \Rightarrow 9 + 6f + c = 0$.Since it touches the $Y$-axis at $(0, 3)$,the $Y$-coordinate of the centre must be $3$,so $-f = 3 \Rightarrow f = -3$.Substituting $f = -3$ into $9 + 6f + c = 0$,we get $9 - 18 + c = 0 \Rightarrow c = 9$.The length of the intercept on the $X$-axis is $2\sqrt{g^2 - c} = 2$,so $g^2 - c = 1$.Substituting $c = 9$,we get $g^2 - 9 = 1$ $\Rightarrow g^2 = 10$ $\Rightarrow g = \pm \sqrt{10}$.Since the centre is $(\alpha, \beta) = (-g, -f)$ and $\alpha  0$. Thus,$g = \sqrt{10}$.Therefore,$\alpha = -g = -\sqrt{10}$ and $\beta = -f = 3$.The centre is $(-\sqrt{10}, 3)$.

If $C(\alpha, \beta)$ with $\alpha < 0$ is the centre of the circle that touches the $Y$-axis at $(0, 3)$ and makes an intercept of length $2$ units on the positive $X$-axis,then $(\alpha, \beta) =$

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