Circles
Circle
Allen Star Batch
Grade 11
Question:
If the conics equations are $S \equiv \sin^2 \theta x^2 + 2h \tan \theta xy + \cos^2 \theta y^2 + 32x + 16y + 19 = 0$, $S' \equiv \cos^2 \theta x^2 + 2h' \cot \theta xy + \sin^2 \theta y^2 + 16x + 32y + 19 = 0$ intersect at four concyclic points, then: (where $\theta \in [0, \pi/2]$)
$h + h' = 0$
$h - h' = 0$
$\theta = \pi/4$
None of these
Step-by-Step Solution
Key Concept: For the radical axis (or intersection curve) to represent a circle, coefficients of $x^2$ and $y^2$ must be equal and the $xy$ coefficient must vanish.
The curve through the intersection of $S_1$ and $S_2$ is given by $S_1 - \lambda S_2 = 0$. Expanding this yields $x^2(\sin^2\theta + \lambda\cos^2\theta) + 2(0)(\tan\theta + 2h'\cot\theta)xy - (\cos^2\theta + \lambda\sin^2\theta)y^2 + (32+16\lambda)x + (16+32\lambda)y + 19(1+\lambda) = 0$. For this to represent a circle, the coefficient of $xy$ must be zero (already satisfied) and the coefficients of $x^2$ and $y^2$ must be equal: $(1-\lambda)\sin^2\theta = (1-\lambda)\cos^2\theta$, which gives $\lambda = 1$ or $\theta = \pi/4$. The condition $h\tan\theta - \lambda h'\cot\theta = 0$ is satisfied when $\lambda = -1$, $\theta = \pi/4$, and $h + h' = 0$.
Correct Answer: 1,2,3