The value of $\lambda$,for which the circle $x^2 + y^2 + 2\lambda x + 6y + 1 = 0$ intersects the circle $x^2 + y^2 + 4x + 2y = 0$ orthogonally is
- A$\frac{-5}{2}$
- B$-1$
- C$\frac{-11}{8}$
- D$\frac{-5}{4}$
Explore More
Similar Questions
The centres of all circles passing through the points of intersection of the circles $x^2+y^2+2x-2y+1=0$ and $x^2+y^2-2x+2y-2=0$ and having radius $\sqrt{14}$ lie on the curve
Where do the circles $x^2 + y^2 + 2x - 2y + 1 = 0$ and $x^2 + y^2 - 2x - 2y + 1 = 0$ touch each other?
Difficult
View SolutionIf the circles $x^2 + y^2 - 16x - 20y + 164 = r^2$ and $(x - 4)^2 + (y - 7)^2 = 36$ intersect at two distinct points,then
DifficultJEE MAIN 2019
View SolutionThe gradient of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $3{x^2} + 3{y^2} - 7x + 8y + 11 = 0$ is
Medium
View SolutionFind the equation of a circle which cuts the circle $x^2+y^2-6x+4y-3=0$ orthogonally,while passing through $(3,0)$ and touching the $Y$-axis.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo