Let the set of all values of r , for which the circles $(x + 1)^{2} + (y + 4)^{2} = r^{2}$ and $x^{2

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Let the set of all values of r , for which the circles $(x + 1)^{2} + (y + 4)^{2} = r^{2}$ and $x^{2} + y^{2} - 4x - 2y - 4 = 0$ intersect at two distinct points be the interval ( $\alpha$, $\beta$ ). Then $\alpha\beta$ is equal to

Accepted Answer

$(x - 2)^{2} + (y - 1)^{2} = 3^{2}\&(x + 1)^{2} + (y + 4)^{2} = r^{2}$

$$\begin{matrix} & \left| r_{1} - r_{2} \right| < c_{1}c_{2} < r_{1} + r_{2} \\ & \ |r - 3| < \sqrt{(2 + 1)^{2} + (1 + 4)^{2}} < r + 3 \\ & \ |r - 3| < \sqrt{34}\& r + 3 > \sqrt{34} \\ & \ - \sqrt{34} < r - 3 < \sqrt{34}\& r > \sqrt{34} - 3 \end{matrix}$$

i.e. $r = (3 - \sqrt{34},3 + \sqrt{34}) \cap (\sqrt{34} - 3,\infty)$

i.e. $r \in (\sqrt{34} - 3,\sqrt{34} + 3)$

$$\begin{matrix} & \ \therefore\alpha\beta = (\sqrt{34} - 3)(\sqrt{34} + 3) \\ & \ = 34 - 9 \\ & \ = 25 \end{matrix}$$

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