EllipseMedium
Question
If a > 2b > 0 then positive value of m for which y = mx - b
is a common tangent tox2 + y2 = b2 and (x - a)2 + y2 = b2 is
is a common tangent tox2 + y2 = b2 and (x - a)2 + y2 = b2 is Options
A.

B.

C.

D.

Solution
Given, y = mx - b
touches both the circles, so distance from centre = radius of both the circles.

and
⇒
⇒ m2a22 - 2abm
b2 (1+ m2) = b2 (1 + m2)
⇒ ma - 2b
= 0
⇒ m2a2 = 4b2 (1 + m2)
⇒
touches both the circles, so distance from centre = radius of both the circles.
and

⇒

⇒ m2a22 - 2abm
b2 (1+ m2) = b2 (1 + m2)⇒ ma - 2b
= 0⇒ m2a2 = 4b2 (1 + m2)
⇒

Create a free account to view solution
View Solution FreeMore Ellipse Questions
If the tangent at the point $\left( 4cos\theta,\frac{16sin\theta}{\sqrt{11}} \right)$ to ellipse $16x^{2} + 11y^{2} = 25...The eccentricity of the ellipse represented by the equation 25x2 + 16y2 − 150x − 175 = 0 is -...The angle between the tangents drawn from the point $(\sqrt{7},1)$ to the ellipse $3x^{2} + 5y^{2} = 15$ is :...Maximum length of chord of the ellipse = 1, such that eccentric angles of its extremities differ by is -...If the points of intersection of the ellipses $x^{2} + 2y^{2} - 6x - 12y + 23 = 0$ and $4x + 2y^{2} - 20x - 12y + 35 = 0...