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

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The number of common tangents on them is 2

They cut orthogonal

They have a common tangent of length $5 {(\frac{12}{5})}^{1 / 4}$

They have a common chord of length $5 \sqrt{\frac{3}{2}}$

answer is A, B, C, D.

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Detailed Solution

$r_{1} = 5, r_{2} = \sqrt{15} c_{1} c_{2} = \sqrt{40}$

$a) | r_{1} - r_{2} | < c_{1} c_{2} < r_{2} + r_{2} ∴$ they intersect

b) 2(1) (3) + 2(2) (–4) = –20 + 10 $∴$ cut orthogonal

c) Common tangent $= \sqrt{d^{2} - {(r_{1} - r_{2})}^{2}} = 5 {(\frac{12}{5})}^{1 / 4}$

d) Common chord $\frac{2 r_{1} r_{2}}{\sqrt{{r_{1}}^{2} + {r_{2}}^{2}}} = 5 \sqrt{\frac{3}{2}}$

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