JEE Main 2025 (Online) 8th April Evening ShiftQuestion 1
Let $r$ be the radius of the circle, which touches $x$-axis at point $(a,0)$, $a<0$ and the parabola $y^{2}=9x$ at the point $(4,6)$. Then $r$ is equal to $\_\_\_\_\_\_$ .
Parabola
Answer$30$
Solution
Concept. The centre lies on the normal at the point of contact, and touching the $x$-axis at $(a,0)$ means the centre is $(a,r)$. Formulas used. (1) $y^2=9x$: slope of tangent $\dfrac{dy}{dx}=\dfrac{9}{2y}$.(2) Radius $=$ distance from the centre to the point of contact.Step 1 - Tangent and normal at (4, 6). Tangent slope $=\dfrac{9}{12}=\dfrac34$, normal slope $=-\dfrac43$. Step 2 - Centre. Centre $(a,r)$ is on the normal: $r-6=-\dfrac43(a-4)$, i.e. $a-4=-\dfrac34(r-6)$. Step 3 - Use (2). $(a-4)^2+(r-6)^2=r^2$: $$\frac{9}{16}(r-6)^2+(r-6)^2=r^2\Rightarrow\frac{25}{16}(r-6)^2=r^2.$$ Step 4 - Solve. $5(r-6)=\pm4r$, so $r=30$ or $r=\dfrac{10}{3}$. Step 5 - Check $a<0$. $r=30$: $a=4-\frac34\cdot24=-14<0$ (valid). $r=\frac{10}3$: $a=4+2=6>0$ (rejected).Answer: 30