Question

P is a point $(a,b)$ in the first quadrant. If the two circles which pass through P and touch both the co-ordinates axes cut at right angles, then

• A. $a^2-6ab+b^2=0$
• B. $a^2+2ab-b^2=0$
• C. $a^2-4ab+b^2=0$
• D. $a^2-8ab+b^2=0$

Answer 1

The correct option is C $a^2-4ab+b^2=0$

Equation of the two circles be $(x-r{)}^2+(y-r{)}^2=r^2$ (since it touches both the axes, X coordinate of center = Y coordinate of center = r)
i.e. $x^2+y^2-2rx-2ry+r^2=0$, where $r=r_1$ and $r_2$. Condition of orthogonality gives
$2r_1{r}_2+2r_1{r}_2=r_1^2+r_2^2\Rightarrow 4r_1{r}_2=r_1^2+r_1^2$.....(1)
Circle passes through (a, b)
$\Rightarrow a^2+b^2-2ra-2rb+r^2=0$
i.e. $r^2-2r(a+b)+a^2+b^2=0$
$r_1+r_2=2(a+b)$ and $r_1{r}_2=a^2+b^2$ ($r_1,r_2$ are roots of above equation)
Using these values in (1), we get
$\therefore 4(a^2+b^2)=4(a+b{)}^2-2(a^2+b^2)$
i.e. $a^2-4ab+b^2=0$