The slope of one of the direct common tangent | vedclass

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

Question

(B) Given circles are $C_1: x^2+y^2-2x+4y+1=0$ and $C_2: x^2+y^2-4x-2y+4=0$. Centers are $C_1(1, -2)$ and $C_2(2, 1)$. Radii are $r_1 = \sqrt{1^2+(-2)

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(B) Given circles are $C_1: x^2+y^2-2x+4y+1=0$ and $C_2: x^2+y^2-4x-2y+4=0$. Centers are $C_1(1, -2)$ and $C_2(2, 1)$. Radii are $r_1 = \sqrt{1^2+(-2)^2-1} = 2$ and $r_2 = \sqrt{2^2+1^2-4} = 1$. Let the common tangent be $y-mx-c=0$. The distance from center to the tangent equals the radius: $\frac{|-2-m(1)-c|}{\sqrt{1+m^2}} = 2$ and $\frac{|1-m(2)-c|}{\sqrt{1+m^2}} = 1$. For direct common tangents,the centers lie on the same side,so $\frac{-2-m-c}{\sqrt{1+m^2}} = 2$ and $\frac{1-2m-c}{\sqrt{1+m^2}} = 1$. Subtracting the equations: $\frac{-3+m}{\sqrt{1+m^2}} = 1$. Squaring both sides: $m^2-6m+9 = 1+m^2$,which gives $6m = 8$,so $m = \frac{4}{3}$.

The slope of one of the direct common tangents drawn to the circles $x^2+y^2-2x+4y+1=0$ and $x^2+y^2-4x-2y+4=0$ is

Similar Questions

Two tangents are drawn from a point $P$ on the radical axis to two circles,touching them at $Q$ and $R$ respectively. Then,the triangle formed by joining $P, Q,$ and $R$ is:

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k$ is equal to :

One of the limit points of the coaxial system of circles containing $x^2 + y^2 - 6x - 6y + 4 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$ is:

If the circles $x^2+y^2-8x-8y+28=0$ and $x^2+y^2-8x-6y+25-\alpha^2=0$ have only one common tangent,then $\alpha=$

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ lies on the line

Vedclass Test Series

Exam Paper Generator

Online Exam Module