Question

If the circles $(x+1{)}^2+(y-1{)}^2=4a^2$ and $x^2+y^2-4x+6y-3=0$ have three common tangents only, then the value of the expression $a^2-2a+\frac{3}{4}$, is

• A. $1$
• B. $0$
• C. $64$
• D. $-1$

Answer 1

The correct option is B $0$

$S_1:(x+1{)}^2+(y-1{)}^2=4a^2\Rightarrow C_1\equiv (-1,1)$
$S_2:x^2+y^2-4x+6y-3=0\Rightarrow C_2\equiv (2,-3)$
$S_1$ & $S_2$ have three common tangents, then $S_1$ & $S_2$ are touching externally each other.
$\therefore r_1+r_2=C_1{C}_1$
$\Rightarrow 2a+\sqrt{2^2+3^2+3}=\sqrt{3^2+4^2}$
$\Rightarrow 2a+\sqrt{16}=5$
$\Rightarrow 2a+4=5$
$\Rightarrow a=\frac{1}{2}$
$\therefore a^2-2a+\frac{3}{4}=\frac{1}{4}-1+\frac{3}{4}=0$

Hence, option B.