

∵316+α=6 and 315+β=6⇒(α,β)≡(2,3) Also, C1C2=r1+r2 ⇒(2−8)2+(3−215)2=2r2+r2⇒r2=25⇒r1=2r2=5∴(α+β)+4(r12+r22)=5+4(425+25)=130
JEE Main 2024 — Mathematics Coordinate Geometry
Let the circles C1:(x−α)2+(y−β)2=r12 and C2:(x−8)2+(y−215)2=r22 touch each other externally at the point (6,6). If the point (6,6) divides the line segment joining the centres of the circles C1 and C2 internally in the ratio 2:1, then (α+β)+4(r12+r22) equals
Held on 8 Apr 2024 · Verified 6 Jul 2026.
125
130
110
145
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Suppose that two chords, drawn from the point $(1, 2)$ on the circle $x^2 + y^2 + x - 3y = 0$ are bisected by the $y$-axis. If the other ends of these chords are $R$ and $S$, and the mid point of the line segment $RS$ is $(\alpha, \beta)$, then $6(\alpha + \beta)$ is equal to:
Let $\dfrac{x^2}{f(a^2+7a+3)} + \dfrac{y^2}{f(3a+15)} = 1$ represent an ellipse with major axis along $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbb{R}$. If the set of all possible values of $a$ is $\mathbb{R} - [\alpha, \beta]$, then $\alpha^2+\beta^2$ is equal to:
Let the vertex $A$ of a triangle $ABC$ be $(1, 2)$, and the mid-point of the side $AB$ be $(5, -1)$. If the centroid of this triangle is $(3, 4)$ and its circumcenter is $(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:
Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the mid point of the line segment PQ is $\left(\dfrac{1}{2}, -\dfrac{1}{2}\right)$. Then the area of the triangle OPQ equals:
If the line $\alpha x+4 y=\sqrt{7}$, where $\alpha \in \mathbf{R}$, touches the ellipse $3 x^{2}+4 y^{2}=1$ at the point P in the first quadrant, then one of the focal distances of $P$ is :
Work through every JEE Main Coordinate Geometry PYQ, year by year.