The radical axis of the co-axial system of ci | vedclass

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

Question

(D) The radical axis of a co-axial system of circles with limiting points $A(1, 2)$ and $B(-2, 1)$ is the perpendicular bisector of the line segment $

Vedclass · Accepted Answer

$3x + y = 0$

Vedclass · Answer

(D) The radical axis of a co-axial system of circles with limiting points $A(1, 2)$ and $B(-2, 1)$ is the perpendicular bisector of the line segment $AB$.First,find the slope of the line segment $AB$:$m_{AB} = \frac{1 - 2}{-2 - 1} = \frac{-1}{-3} = \frac{1}{3}$.The slope of the perpendicular bisector is $m_{\perp} = -\frac{1}{m_{AB}} = -3$.The midpoint of $AB$ is $M = \left(\frac{1 - 2}{2}, \frac{2 + 1}{2}\right) = \left(-\frac{1}{2}, \frac{3}{2}\right)$.The equation of the radical axis is $y - y_1 = m_{\perp}(x - x_1)$:$y - \frac{3}{2} = -3(x + \frac{1}{2})$$y - \frac{3}{2} = -3x - \frac{3}{2}$$3x + y = 0$.Thus,option $D$ is correct.

The radical axis of the co-axial system of circles with limiting points $(1, 2)$ and $(-2, 1)$ is

Similar Questions

The line $x-2=0$ cuts the circle $x^2+y^2-8x-2y+8=0$ at $A$ and $B$. The equation of the circle passing through the points $A$ and $B$ and having the least radius is

If $(\alpha, \beta)$ is the centre of the circle which passes through the point $(1, -1)$ and cuts the circles $x^2+y^2+2x-3y-5=0$ and $x^2+y^2-3x+2y+1=0$ orthogonally,then $\alpha-5\beta=$

If the equation of the common tangent of the circles $x^2+y^2-4x+6y+4=0$ and $x^2+y^2+2x-2y-2=0$ at their point of contact is $ax+by+c=0$,then $\frac{a}{c}=$

Find 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.

The point of intersection of the common tangents drawn to the circles $x^2+y^2-4x-2y+1=0$ and $x^2+y^2-6x-4y+4=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module