The radical centre of three circles described | vedclass

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

Question

(A) Let the vertices of the triangle be $A, B,$ and $C$. Consider the three circles having the sides $AB, BC,$ and $CA$ as their diameters.The equatio

Vedclass · Accepted Answer

The orthocentre

Vedclass · Answer

(A) Let the vertices of the triangle be $A, B,$ and $C$. Consider the three circles having the sides $AB, BC,$ and $CA$ as their diameters.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.The radical axis of any two of these circles is the common chord of the two circles. For circles described on the sides of a triangle as diameters,the radical axis of any two circles is the altitude of the triangle passing through the vertex common to the two sides.Since the radical centre is the point of intersection of the radical axes of the three circles taken in pairs,it is the point of intersection of the three altitudes of the triangle.Therefore,the radical centre is the orthocentre of the triangle.

The radical centre of three circles described on the three sides of a triangle as diameters is

Similar Questions

The equation of the circle passing through the points of intersection of $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 4y + 1 = 0$ and touching the line $x + 2y = 0$ is

If one of the diameters of the circle $x^{2}+y^{2}-2x-6y+6=0$ is a chord of another circle $'C'$,whose center is at $(2,1)$,then its radius is..........

The circles $x^2+y^2+2x+3y-7=0$ and $x^2+y^2+4x-7y+5=0$ intersect at the points $A$ and $B$. The equation of the circle,having $\overline{AB}$ as a diameter is

The radical axis of any two circles is $ \dots $ to the line joining their centres.

The number of common tangents to the two circles $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module