Question

Let P(a,b) be a point in the first quadrant. Two circles are drawn through P touching the co-ordinate axes. If one circle bisects the circumference of other, then possible values of $\frac{a}{b}$ - is/are-

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

Answer 1

Answer: (A)

a² - 6ab + b² = 0

Answer 2

The equation of a circle touching the coordinate axes in the first quadrant with radius $r$ is $(x-r)^2 + (y-r)^2 = r^2$. If P(a,b) lies on this circle, then $(a-r)^2 + (b-r)^2 = r^2$, which simplifies to $r^2 - 2(a+b)r + (a^2+b^2) = 0$. Let the radii of the two circles be $r_1$ and $r_2$. Then $r_1$ and $r_2$ are the roots of this quadratic equation. By Vieta's formulas, $r_1 + r_2 = 2(a+b)$ and $r_1 r_2 = a^2+b^2$.

The condition that one circle bisects the circumference of the other means that the common chord of the two circles is a diameter of one of them. The equation of the common chord is obtained by subtracting the equations of the two circles: $(x^2 + y^2 - 2r_1x - 2r_1y + r_1^2) - (x^2 + y^2 - 2r_2x - 2r_2y + r_2^2) = 0$ $2(r_2-r_1)x + 2(r_2-r_1)y + (r_1^2 - r_2^2) = 0$ Assuming $r_1 \neq r_2$, we divide by $r_2-r_1$: $2x + 2y - (r_1+r_2) = 0$

If the common chord is a diameter of the circle with radius $r_2$, its center $(r_2, r_2)$ must lie on the common chord: $2r_2 + 2r_2 - (r_1+r_2) = 0$ $4r_2 - r_1 - r_2 = 0$ $3r_2 = r_1$

This implies that one radius is three times the other. Let the radii be $r$ and $3r$. Thus, $\{r_1, r_2\} = \{r, 3r\}$. Using Vieta's formulas: Sum of radii: $r_1 + r_2 = r + 3r = 4r$. From Vieta's formulas, $r_1 + r_2 = 2(a+b)$. So, $4r = 2(a+b) \implies 2r = a+b$. Product of radii: $r_1 r_2 = r \cdot 3r = 3r^2$. From Vieta's formulas, $r_1 r_2 = a^2+b^2$. So, $3r^2 = a^2+b^2$.

Substitute $r = \frac{a+b}{2}$ into the product equation: $3 \left(\frac{a+b}{2}\right)^2 = a^2+b^2$ $3 \frac{(a+b)^2}{4} = a^2+b^2$ $3(a^2 + 2ab + b^2) = 4(a^2+b^2)$ $3a^2 + 6ab + 3b^2 = 4a^2 + 4b^2$ $a^2 - 6ab + b^2 = 0$