Question

There are two circles whose equations are $x ^ { 2 } + y ^ { 2 } = 9$ and If the two circles have exactly two common tangents, then the number of possible values of n is

• A. 2
• B. 8
• C. 9
• D. None of these

Answer 1

Answer: (C)

9

Answer 2

For $x ^ { 2 } + y ^ { 2 } = 9$, the centre = (0, 0) and the radius = 3

For $x ^ { 2 } + y ^ { 2 } - 8 x - 6 y + n ^ { 2 } = 0$.

The centre = (4, 3) and the radius $= \sqrt { ( 4 ) ^ { 2 } + ( 3 ) ^ { 2 } - n ^ { 2 } }$

$\therefore$ $4 ^ { 2 } + 3 ^ { 2 } - n ^ { 2 } > 0$ or $n ^ { 2 } < 5 ^ { 2 }$ or

Circles should cut to have exactly two common tangents.

So, $\therefore$ $3 + \sqrt { 25 - n ^ { 2 } } > \sqrt { ( 4 ) ^ { 2 } + ( 3 ) ^ { 2 } }$ or

$\sqrt { 25 - n ^ { 2 } } > 2$ or $25 - n ^ { 2 } > 4$

$\therefore$ $n ^ { 2 } < 21$ or $- \sqrt { 21 } < n < \sqrt { 21 }$

Therefore, common values of n should satisfy

$- \sqrt { 21 } < n < \sqrt { 21 }$

But $n \in Z$, So, .

$\therefore$ Number of possible values of n = 9