The circles $x^{2} + y^{2} + 2 x + 4 y - 20 = 0 and x^{2} + y^{2} + 6 x - 8 y + 10 = 0$

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are such that the number of common tangents on them is 2

are orthogonal

are such that the length of their common tangent is $5 (12 / 5)^{1 / 4}$

$are such that the length of their common chord is 5 \sqrt{3 / 2}$

answer is A, B, C, D.

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Detailed Solution

$r_{1} = 5; r_{2} = \sqrt{15}; C_{1} C_{2} = \sqrt{40} ∴ r_{1} + r_{2} > C_{1} C_{2} > r_{1} - r_{2} Hence, the circles intersect at two distinct points . There are two common tangents . Also, 2 g_{1} g_{2} + 2 f_{1} f_{2} = 2 (1) (3) + 2 (2) (- 4) = - 10 and c_{1} + c_{2} = - 20 + 10 = - 10 Thus, the two circles are orthogonal Length of common chord = \frac{2 r_{1} r_{2}}{\sqrt{r_{1}^{2} + r_{2}^{2}}} = 5 \sqrt{\frac{3}{2}} Length of common tangent = \sqrt{C_{1} C_{2}^{2} - {(r_{1} - r_{2})}^{2}} = 5 {(\frac{12}{5})}^{1 / 4} .$