If the circles x^2+y^2=9 and x^2+y^2+8 y+c=0 touch each other, th… - Vidyadip

MCQ

If the circles $x^2+y^2=9$ and $x^2+y^2+8 y+c=0$ touch each other, then c is equal to:

✓
15
B
-15
C
16
D
-16

✓

Answer

Correct option: A.

15

15

Solution:
The centre of the circle $x^2+y^2=9$ is $(0,0)$.
Let us denote it by $\mathrm{C}_1$.
The centre of the circle $x^2+y^2+8 y+c=0$ is $(0,-4)$.
Let us denote it by $\mathrm{C}_2$.
The radius of $x^2+y^2=9$ is 3 units.
$x^2+y^2+8 y+c=0$
$\Rightarrow(\text{x}-0)^2+(\text{y}+4)^2=16-\text{c}=(\sqrt{16-\text{c}})^2$
Therefore, the radius of the above circle is $\sqrt{16-\text{c}}$
Let the circles touch each other at P.
$\therefore\text{C}_1\text{C}_2=\text{PC}_2+\text{PC}_1$
$\Rightarrow\text{PC}_2=4-3=1$
$\Rightarrow\text{PC}_2-1=\sqrt{16-\text{c}}$
$\Rightarrow\text{c}=15$

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