Circles
Circle
Allen Star Batch
Grade 11
Question:
MATCH THE FOLLOWING:
(A) A circle of constant radius '$a$' passes through origin '$O$' and cuts the axes of coordinates in points $P$ and $Q$, then the equation of the locus of the foot of perpendicular from $O$ to $PQ$ is $(x^2 + y^2)\left(\frac{1}{x^2} + \frac{1}{y^2}\right) = ka^2$, then $k$ is
(B) Tangents are drawn from any point on the circle $x^2 + y^2 = R^2$ to the circle $x^2 + y^2 = r^2$. The line joining the points of intersection of these tangents with circle also touch the second. If $R$ equals $k r$, then $k$ is
(C) A ray of light incident at the point $(-2, -1)$ gets reflected from the tangent at $(0, -1)$ to the circle $x^2 + y^2 = 1$. The reflected ray touches the circle. If the equation of the line along which the incident ray moves is $ay + bx = 11$ then the value of $a + b$ is
(D) The equation of a line inclined at an angle $\frac{5}{4}$ to the axis $X$, such that the two circle $x^2 + y^2 = 4$, $x^2 + y^2 - 10x - 14y + 65 = 0$ intercept equal lengths on it is $ax + by - 3 = 0$, then the value of $a + b$ is
Step-by-Step Solution
Key Concept: The chord of contact, radical axis, and reflection geometry are fundamental properties connecting external points to circles.
For part (A), the equation of chord of contact $PQ$ from external point $P(h,k)$ to circle $x^2 + y^2 = a^2$ is derived using the property that $PQ^2 = 4a^2$ when $(h^2 + k^2)(\frac{1}{h^2} + \frac{1}{k^2}) = 4a^2$. For part (B), using $r = R\cos 60° = \frac{R}{2}$ relates the inradius to circumradius. Part (C) calculates $\tan 2\theta = \frac{4}{3}$ leading to the incident ray equation $3y + 4x + 11 = 0$. Part (D) uses the radical axis condition $r_1^2 - P_1^2 = r_2^2 - P_2^2$ to find $C = -\frac{3}{2}$, giving the line equation $y = x - \frac{3}{2}$.
Correct Answer: [A-r, B-s, C-p, D-q]