Question

The internal common tangents of the circles $x^2+y^2-4x-4y+4=0$ and $x^2+y^2+6x+6y+9=0$ are

• A. $x=0$
• B. $x-y=2$
• C. $y=0$
• D. $x+y=2$

Answer 1

Answer: (A)

$x=0$

As we have $x^2+y^2-4x-4y=0$

$⟹(x-2{)}^2+(y-2{)}^2=2^2$

Centre $=C_1(2,2)$ with radius $=2$

From another equation of circle

$x^2+y^2+6x+9=0$

$⟹(x+3{)}^2+(y+3{)}^2=3^2$

Centre $=C_2(-3,3)$ with radius $=3$

To find the internal common tangents of circle with the help of section formula

$x=\frac{-3\left(2\right)+2\left(3\right)}{3+2}⟹x=0$