The equation of the radical axis of the circl | vedclass

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

Question

(C) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.First,rewrite the equations in the form $x^2 +

Vedclass · Accepted Answer

$7x - 8y - 14 = 0$

Vedclass · Answer

(C) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.First,rewrite the equations in the form $x^2 + y^2 + 2gx + 2fy + c = 0$ with coefficient $1$ for $x^2$ and $y^2$.For the first circle: $2x^2 + 2y^2 - 7x = 0 \Rightarrow x^2 + y^2 - 3.5x = 0$.For the second circle: $x^2 + y^2 - 4y - 7 = 0$.Subtracting the two equations: $(x^2 + y^2 - 3.5x) - (x^2 + y^2 - 4y - 7) = 0$.$-3.5x + 4y + 7 = 0$.Multiply by $-2$ to clear the decimal: $7x - 8y - 14 = 0$.

The equation of the radical axis of the circles $2x^2 + 2y^2 - 7x = 0$ and $x^2 + y^2 - 4y - 7 = 0$ is

Similar Questions

Find the equation of the circle whose diameter is the common chord of the circles $x^2 + y^2 - 8x + y - 15 = 0$ and $x^2 + y^2 - 4x + 4y - 42 = 0$.

If one of the diameters of the circle,given by the equation $x^2 + y^2 - 4x + 6y - 12 = 0$,is a chord of a circle $S$ whose center is at $(-3, 2)$,then the radius of $S$ 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 :

If the circle $x^2 + y^2 + 4x + 22y + c = 0$ bisects the circumference of the circle $x^2 + y^2 - 2x + 8y - d = 0$,then $c + d = . . . . .$

The coordinates of the point from where the tangents drawn to the circles ${x^2} + {y^2} = 1$,${x^2} + {y^2} + 8x + 15 = 0$,and ${x^2} + {y^2} + 10y + 24 = 0$ are of the same length are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module