Question

Equation of a circle with centre $(4,3)$ touching the circle $x^2+y^2=1$ is

• A. $x^2+y^2-8x-6y+9=0$
• B. $x^2+y^2+8x+6y-11=0$
• C. $x^2+y^2-8x-6y-11=0$
• D. $x^2+y^2+8x+6y-9=0$

Answer 1

Answer: (C), (D)

The correct options are
C $x^2+y^2-8x-6y-11=0$
D $x^2+y^2+8x+6y-9=0$

Let equation of the required circle be $(x-4{)}^2+(y-3{)}^2=r^2\left(1\right)$

If the circle $\left(1\right)$ touches the circle $x^2+y^2=1$, then distance

between the centres $(4,3)$ and $(0,0)$ of these circles is equal to the

sum or difference of their radii, $r$ and $1$

$\Rightarrow \sqrt{(4^2-0)+(3^2-0)}=1+r$

$\Rightarrow r+1=5$

$\Rightarrow r=4$

so the equations of the required circles from $\left(1\right)$ are

$(x-4{)}^2+(y-3{)}^2=r^2$

$x^2-8x+16+y^2-6y+9=16$

$x^2+y^2-8x-6y+9=0$