If the circles x^2 + y^2 = 9 and x^2 + y^2 + 8y + c = 0 touch eac… - Vidyadip

MCQ

If the circles $x^2 + y^2 = 9$ and $x^2 + y^2 + 8y + c = 0$ 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 $C_1$.
The centre of the circle $x^2 + y^2+ 8y + c = 0$ is $(0, -4)$.
Let us denote it by $C_2$.
The radius of $x^2 + y^2 = 9$ is $3$ units.
$x^2 + y^2+ 8y + c = 0$
$\Rightarrow(\text{x}-0)^2+(\text{y}+4)^2=16-\text{c}=(\sqrt{16-\text{c}})^2$
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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