Question

The number of common tangents to the circles $x^2+y^2=4$ and $x^2+y^2-6x-8y-24=0$ is

• A. 4
• B. 3
• C. 1
• D. 2

Answer 1

Answer: (A)

4

Answer 2

Given, equation of circles are
$x^2 + y^2 = 4$,
whose radius = 2,
centre = (0, 0)
and $x^2 + y^2 - 6 x - 8 y - 24 = 0,$
whose radius $= \sqrt{9 + 16- 24} = \sqrt{1} = 1$
and centre = (3 4)
Now, $c_1 \,c_2 = \sqrt{(3 - 0)^2 + (4 - 0)^2}$
$= \sqrt{ 9 + 16} = 5$
$a_1 + a_2 = 2 + 1 = 3$
Since, $c_1 c_2 > a_1 + a_2$
$\therefore$ Number of common tangents = 4