Question

A circle $C_1$ of unit radius touches coordinate axes in first quadrant. Another circle $C_2$ touches $C_1$ externally and also touches each line of pair $xy-4x-4y+16=0$. Then radius $C_2$ is

• A. $4\sqrt{2}-3$
• B. $4-3\sqrt{2}$
• C. $7-4\sqrt{2}$
• D. $6-4\sqrt{2}$

Answer 1

Answer: (C)

$7-4\sqrt{2}$

$Fromulaused$
$D\text{istance between two point A}(x_1,y_1)andB(x_2,y_2)is$
$AB=\sqrt{{(y_2-y_1)}^2+{(x_2-x_1)}^2}$
$Let{\text{the radius of circle C}}_2=r$
$Radiusofcircle{C}_1=1\left(given\right)$
$xy-4x-4y+16=0$
$\Rightarrow x(y-4)-4(y-4)=0$
$\Rightarrow (y-4)(x-4)=0$
$Hencelinesarex=4,y=4$
$Nowconsider$
${\text{The diameter of circle C}}_{1}and{C}_2\text{and the lines x=4 and y=4 form a straight line}$
$Hence$
$O{C}_1+C_1{C}_2+C_{2}A=OA$
$\Rightarrow \sqrt{1^2+1^2}+1+r+\sqrt{r^2+r^2}=\sqrt{4^2+4^2}$
$\Rightarrow \sqrt{2}+1+r+r\sqrt{2}=4\sqrt{2}$
$\Rightarrow 1+r+r\sqrt{2}=3\sqrt{2}$
$\Rightarrow r(1+\sqrt{2})=3\sqrt{2}-1$
$\Rightarrow r=\frac{3\sqrt{2}-1}{1+\sqrt{2}}$
$rationalizing$
$r=\frac{3\sqrt{2}-1}{1+\sqrt{2}}\times \frac{1-\sqrt{2}}{1-\sqrt{2}}$
$=7-4\sqrt{2}$