Circles
Circle
Allen Star Batch
Grade 11
Question:
If two points $A(-2, a)$ and $B(4, \beta)$ are such that the triangle $AOB$ is the right-angle triangle right angle at $O$. If $S = 0$ be the equation of the locus of the foot of the perpendicular $P$ drawn from the point $O$ from the line $AB$, then:
$(1, 3)$ lies on $S = 0$
Minimum possible area of triangle $AOB$ is $9$
Locus is a parabola
Locus is a circle
Step-by-Step Solution
Key Concept: A variable family of circles passing through two fixed curves always passes through their points of intersection and one additional fixed point.
The locus is determined by substituting $b = a\left(x - \frac{m}{n}\right)y$ into the circle equation $x^2 + y^2 - \frac{1}{n}y = 0$, which gives $\left(x^2 + y^2 - \frac{1}{n}y\right) - a\left(x - \frac{m}{n}y\right) = 0$ for all values of $a$. This represents a family of circles that all pass through the intersection points of $x^2 + y^2 - \frac{1}{n}y = 0$ and $x - \frac{m}{n}y = 0$, and the circle always passes through the fixed point $\left(\frac{m}{m^2+n^2}, \frac{n}{m^2+n^2}\right)$.
Correct Answer: 1,4