Question

If the circle x2 + y2 = 9 touches the circle x2 + y2 + 6y + c = 0, then sum of all the values of c is equal to​

Answer 1

Answer:

A circle

$C_1= {x}^{2} + {y}^{2} = 9$

touches the another circle

$C_2 = {x}^{2} + {y}^{2} + 6y+ c = 0$

Now,

Central first circle will be

$O_1(0.0)$

And its radius will be

3 units.

Also,

centre of second circle

$O_2( - 6.0)$

And radius,

$r_2 = \sqrt{ {6}^{2} + {0}^{2} - c } \\ = \sqrt{36 - c}$

As both touches each other

So,

$C_1C_2 = r_1 + r_2 \\ \sqrt{36} = 3 \: + \sqrt{36 \: - \: c }$

$6 = \: 3 + \sqrt{36 - c} \\ \sqrt{36 - c} = 3 \\ 36 - c = 9 \\ - c = 9 - 36 \\ \displaystyle{\bold{\red{\boxed{\boxed{c = 27}}}}}$

Hence,

for C = 27 both circles touches each other .