MATCH THE FOLLOWING:
(A) $P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the coordinate axes cut at right angles, then the relation between $a$ and $b$ is given by $a^2 + b^2 - kab = 0$, then the value of $k$ is
Step-by-Step Solution
Key Concept: Orthogonality of circles and the discriminant condition ensure existence of real solutions for intersection points.
For part (A), two circles are orthogonal when $2rr_1 + 2rr_2 = r^2 + r_1^2$, leading to $6rr_2 = (r_1 + r_2)^2$. The condition that a circle passes through point $(a,b)$ gives $r^2 - 2(a+b)r + (a^2 + b^2) = 0$. Combining these yields $a^2 + b^2 - 4ab = 0$. Part (B) substitutes $h = \frac{b^2}{2}$ into the circle equation to get $h^2 - ah + \frac{3b^2}{4} = 0$ requiring discriminant $D > 0$ for two distinct roots, giving $a^2 - 3b^2 > 0$. Part (C) uses the geometry that $AE = 2(AB)\cos 60°$. Part (D) applies the perpendicular distance formula between parallel tangents to get $2r = \frac{15}{2\sqrt{5}} = \frac{3}{2}$.
Correct Answer: [A-r, B-p, C-s, D-p]