The equations of the circles are x2+y2−10x−10y+λ=0 and x2+y2−4x−4y+6=0 C1= centre of (1)=(5,5)C2= centre of (2)=(2,2)d= distance between centres =C1C2=9+9=18r1=50−λ,r2=2 For exactly two common tangents we have r1−r2<C1C2<r1+r2⇒50−λ−2<32<50−λ+2⇒50−λ−2<32 or 32<50−λ+2⇒50−λ<42 or 22<50−λ⇒50−λ<32 or 8<50−λ⇒λ>18 or λ<42 Required interval is (18,42)
The set of all real values of λ for which exactly two common tangents can be drawn to the circles x^2+y^2-4 x-4 y+6=0 and x^2+ y^2-10 x-10 y+λ=0 is…
JEE Main 2014 — Mathematics Coordinate Geometry
The set of all real values of λ for which exactly two common tangents can be drawn to the circles x2+y2−4x−4y+6=0 and x2+y2−10x−10y+λ=0 is the interval:
Held on 11 Apr 2014 · Verified 6 Jul 2026.
Options
- A
(12,32)
- B
(18,42)
- C
(12,24)
- D
(18,48)
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Coordinate Geometry in other years
More JEE Main Coordinate Geometry questions
Let $A,B$ be points on the two half-lines $x-\sqrt{3}|y|=\alpha$, $\alpha>0$ at a distance of $\alpha$ from their point of intersection $P$. The line segment $AB$ meets the angle bisector of the given half-lines at the point $Q$. If $PQ=\dfrac{9}{2}$ and $R$ is the radius of the circumcircle of $\triangle PAB$, then $\dfrac{\alpha^2}{R}$ is equal to ______
If a straight line drawn through the point of intersection of the lines $4x+3y-1=0$ and $3x+4y-1=0$, meets the co-ordinate axes at the points P and Q, then the locus of the mid point of PQ is:
Let $C$ be a circle having centre in the first quadrant and touching the $x$-axis at a distance of $3$ units from the origin. If the circle $C$ has an intercept of length $6\sqrt{3}$ on $y$-axis, then the length of the chord of the circle $C$ on the line $x - y = 3$ is :
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:
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:
Other Mathematics topics for JEE Main
Keep practicing Coordinate Geometry
Work through every JEE Main Coordinate Geometry PYQ, year by year.