The number of common tangents to the circles $ x^{2}+y^{2}+2 x+8 y-23=0 $ and $ x^{2}+y^{2}-4 x-10 y+9=0 $, are

Question

The number of common tangents to the circles $ x^{2}+y^{2}+2 x+8 y-23=0 $ and $ x^{2}+y^{2}-4 x-10 y+9=0 $, are,

MARKS App · Accepted Answer

The centres and radii of given circles are

C_{1} (- 1, - 4), C_{2} (2, 5)

and

r_{1} = \sqrt{1 + 16 + 23} = \sqrt{40}

,

r_{2} = \sqrt{4 + 25 - 9} = \sqrt{20}

Now,

C_{1} C_{2} = \sqrt{{(2 + 1)}^{2} + {(5 + 4)}^{2}} = \sqrt{90}

And

r_{1} + r_{2} = \sqrt{40} + \sqrt{20}

Here,

C_{1} C_{2} < r_{1} + r_{2}

∴

Two common tangents can be drawn