If one common tangent of the two circles x^2 | vedclass

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

Question

(A) The equation of the tangent to the circle $x^2 + y^2 = 4$ at the point $(\sqrt{3}, 1)$ is given by $T = 0$,which is $x x_1 + y y_1 = r^2$.Substitu

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(A) The equation of the tangent to the circle $x^2 + y^2 = 4$ at the point $(\sqrt{3}, 1)$ is given by $T = 0$,which is $x x_1 + y y_1 = r^2$.Substituting the point $(\sqrt{3}, 1)$,we get $\sqrt{3}x + 1y = 4$,or $\sqrt{3}x + y - 4 = 0$.This line is also a tangent to the second circle $x^2 + (y - 3)^2 = \lambda$. The center of this circle is $(0, 3)$ and its radius is $\sqrt{\lambda}$.The perpendicular distance from the center $(0, 3)$ to the line $\sqrt{3}x + y - 4 = 0$ must be equal to the radius $\sqrt{\lambda}$.Distance $d = \frac{|\sqrt{3}(0) + 1(3) - 4|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|3 - 4|}{\sqrt{3 + 1}} = \frac{|-1|}{2} = \frac{1}{2}$.Since $d^2 = \lambda$,we have $\lambda = (\frac{1}{2})^2 = \frac{1}{4}$.

If one common tangent of the two circles $x^2 + y^2 = 4$ and $x^2 + (y - 3)^2 = \lambda, \lambda > 0$ passes through the point $(\sqrt{3}, 1)$,then the possible value of $\lambda$ is

Similar Questions

The normal at a point $\theta$ to the curve $x=a(1+\cos \theta), y=a \sin \theta$ always passes through the fixed point

If $y=3x$ is a tangent to a circle with centre $(1,1)$,then the other tangent drawn through $(0,0)$ to the circle is

The product of the slopes of the common tangents drawn to the circles $x^2+y^2+2x-2y-2=0$ and $x^2+y^2-2x+2y+1=0$ is:

The equation of the normal to the circle $x^2+y^2+6x+4y-3=0$ at $(1,-2)$ is

Find the equation of the normal to the circle $x^2 + y^2 - 5x + 2y - 48 = 0$ at the point $(5, 6)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module