Given that a 2b 0 and that the line y = m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equations of the circles are $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$. Since the line $y = mx - b \sqrt{1 + m^2}$ is tangent to the first

Vedclass · Accepted Answer

$\frac{2b}{\sqrt{a^2 - 4b^2}}$

Vedclass · Answer

(D) The equations of the circles are $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$. Since the line $y = mx - b \sqrt{1 + m^2}$ is tangent to the first circle,it is a standard form for a tangent to $x^2 + y^2 = b^2$. For the second circle $(x - a)^2 + y^2 = b^2$,the tangent line equation is $y = m(x - a) \pm b \sqrt{1 + m^2}$. Since the line is common,we equate the two forms: $mx - b \sqrt{1 + m^2} = mx - ma \pm b \sqrt{1 + m^2}$. Taking the positive sign for the common tangent,we get: $ma = 2b \sqrt{1 + m^2}$. Squaring both sides: $m^2 a^2 = 4b^2 (1 + m^2) = 4b^2 + 4b^2 m^2$. $m^2 (a^2 - 4b^2) = 4b^2$. $m^2 = \frac{4b^2}{a^2 - 4b^2}$. Taking the positive square root: $m = \frac{2b}{\sqrt{a^2 - 4b^2}}$.

Given that $a > 2b > 0$ and that the line $y = mx - b \sqrt{1 + m^2}$ is a common tangent to the circles $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$. Then the positive value of $m$ is

Similar Questions

The equation of the common chord of the circles $x^2+y^2+2x+3y+1=0$ and $x^2+y^2-5x-6y+4=0$ is

The distance between the chords of contact of tangents to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ from the origin and the point $(g, f)$ is:

The length of the common chord of the circles $x^2+y^2-6x-4y+13-c^2=0$ and $x^2+y^2-4x-6y+13-c^2=0$ is

If the line $4x + 4y - 11 = 0$ intersects the circle $x^2 + y^2 - 4x - 6y + 4 = 0$ at $A$ and $B$,then the point of intersection of the tangents drawn at $A$ and $B$ is

For the circles $(x-a)^2+y^2=a^2$ and $x^2+(y-a)^2=a^2$,where $a>0$,which one of the following is not true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module