Straight Lines
Orthocentre — Finding Vertex on Circle
nta_pyq_2024_apr
Grade 11
Question:
If $P(6,1)$ be the orthocentre of the triangle whose vertices are $A(5,-2)$, $B(8,3)$ and $C(h,k)$, then the point $C$ lies on the circle:
$x^2+y^2-61=0$
$x^2+y^2-52=0$
$x^2+y^2-65=0$
$x^2+y^2-74=0$
Step-by-Step Solution
Key Concept: $AP\perp BC$ and $BP\perp AC$. Slope of $AP=\frac{1-(-2)}{6-5}=3$, so slope of $BC=-1/3$. Slope of $BP=\frac{1-3}{6-8}=1$, so slope of $AC=-1$.
$C=(-4,7)$. $C$ lies on $x^2+y^2-65=0$.
Correct Answer: 3