Question

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

• A. a² - 6ab + b² = 0
• B. a² + 2ab - b² = 0
• C. a² - 4ab + b² = 0
• D. a² - 8ab + b² = 0

Answer 1

Answer: (C)

a² - 4ab + b² = 0

Answer 2

Let the equations of the two circles be $(x-r_1)^2 + (y-r_1)^2 = r_1^2$ and $(x-r_2)^2 + (y-r_2)^2 = r_2^2$. Since P(a,b) lies on both circles, we have $(a-r)^2 + (b-r)^2 = r^2$, which simplifies to $r^2 - 2(a+b)r + (a^2+b^2) = 0$. The roots of this quadratic are $r_1$ and $r_2$. From Vieta's formulas, $r_1+r_2 = 2(a+b)$ and $r_1r_2 = a^2+b^2$. The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$. For the two circles, the centers are $(-r_1, -r_1)$ and $(-r_2, -r_2)$, and the radii are $r_1$ and $r_2$. The conditions for the circles are: Circle 1: $g_1 = -r_1$, $f_1 = -r_1$, $c_1 = r_1^2$. Circle 2: $g_2 = -r_2$, $f_2 = -r_2$, $c_2 = r_2^2$. For two circles to intersect at right angles, the condition is $2g_1g_2 + 2f_1f_2 = c_1+c_2$. Substituting the values: $2(-r_1)(-r_2) + 2(-r_1)(-r_2) = r_1^2 + r_2^2$. $4r_1r_2 = r_1^2 + r_2^2$. We know $r_1^2 + r_2^2 = (r_1+r_2)^2 - 2r_1r_2$. So, $4r_1r_2 = (r_1+r_2)^2 - 2r_1r_2$. $6r_1r_2 = (r_1+r_2)^2$. Substitute $r_1+r_2 = 2(a+b)$ and $r_1r_2 = a^2+b^2$: $6(a^2+b^2) = (2(a+b))^2$ $6(a^2+b^2) = 4(a^2+2ab+b^2)$ $3(a^2+b^2) = 2(a^2+2ab+b^2)$ $3a^2 + 3b^2 = 2a^2 + 4ab + 2b^2$ $a^2 - 4ab + b^2 = 0$.