The two circles $x^{2} + y^{2} - 5 = 0$ and $x^{2} + y^{2} - 2x - 4y - 15 = 0$

Question

Dr. Abubacker · Accepted Answer

(b)
Centre and radii of two circles are
$C_{1}(0,\ 0),\ C_{2}(1,\ 2)$ and $r_{1} = \sqrt{5},\ r_{2} = 2\sqrt{5}$
Since, $C_{1}C_{2} = \sqrt{5} = r_{2} - r_{1}$, therefore, the two circles touch each other internally

Q. The two circles $x^{2} + y^{2} - 5 = 0$ and $x^{2} + y^{2} - 2x - 4y - 15 = 0$

Explanation

Reviewed by Dr. Abubacker

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