Question

The number of common tangents of the circles $2x^2+2y^2-2x-2y-7=0$ and $x^2+y^2-4x-8y-5=0$ is

• A. $3$
• B. $2$
• C. $1$
• D. $4$

Answer 1

The correct option is B $2$

The given circles are

$C_1:x^2+y^2-x-y-\frac{7}{2}=0$

$\Rightarrow {(x-\frac{1}{2})}^2+{(y-\frac{1}{2})}^2=4$

$C_1=\left(\frac{1}{2,}\frac{1}{2}\right),r_1=2$
Now, the circle $C_2:x^2+y^2-4x-8y-5=0$

$\Rightarrow (x-2{)}^2+(y-4{)}^2=25$

$C_2=(2,4),r_2=5$
Now, $|r_1-r_2|=3,r_1+r_2=7,C_1{C}_2=\sqrt{\frac{29}{2}}$
$\Rightarrow |r_1-r_2|<C_1{C}_2<r_1+r_2$
$\Rightarrow$ The circles intersect each other at two distinct points.Hence, they have two common tangents.
Hence, option 'B' is correct.