The radical centre of the circles x^2 + y^2 + | vedclass

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

Question

(A) Let the equations of the circles be:$S_1: x^2 + y^2 + 4x + 6y - 19 = 0$$S_2: x^2 + y^2 - 9 = 0$$S_3: x^2 + y^2 - 2x - 2y - 5 = 0$The radical axis

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(A) Let the equations of the circles be:$S_1: x^2 + y^2 + 4x + 6y - 19 = 0$$S_2: x^2 + y^2 - 9 = 0$$S_3: x^2 + y^2 - 2x - 2y - 5 = 0$The radical axis of $S_1$ and $S_2$ is $S_1 - S_2 = 0$:$(x^2 + y^2 + 4x + 6y - 19) - (x^2 + y^2 - 9) = 0$$4x + 6y - 10 = 0 \implies 2x + 3y = 5$ ... $(i)$The radical axis of $S_2$ and $S_3$ is $S_2 - S_3 = 0$:$(x^2 + y^2 - 9) - (x^2 + y^2 - 2x - 2y - 5) = 0$$2x + 2y - 4 = 0 \implies x + y = 2$ ... $(ii)$Solving equations $(i)$ and $(ii)$:From $(ii)$,$y = 2 - x$. Substituting in $(i)$:$2x + 3(2 - x) = 5$$2x + 6 - 3x = 5$$-x = -1 \implies x = 1$Substituting $x = 1$ in $(ii)$,$1 + y = 2 \implies y = 1$.Thus,the radical centre is $(1, 1)$.

The radical centre of the circles $x^2 + y^2 + 4x + 6y = 19$,$x^2 + y^2 = 9$,and $x^2 + y^2 - 2x - 2y = 5$ is:

Similar Questions

The radical axis of circles $x^2+y^2+5x+4y-5=0$ and $x^2+y^2-3x+5y-6=0$ is

If the line $x+y=2$ cuts the circle $x^2+y^2+2x-4y+4=0$ at two points $A$ and $B$,then the radius of the circle passing through $A$ and $B$ and orthogonal to $x^2+y^2-2x-4y-4=0$ is

If $L_1$ represents the radical axis of circles $x^2+y^2-4x-6y+5=0$ and $x^2+y^2-2x-4y-1=0$,and $L_2$ represents the radical axis of $x^2+y^2+2x+2y-7=0$ and $x^2+y^2+x+y+9=0$,then:

The equation of a circle passing through the points of intersection of the circles $x^2 + y^2 + 13x - 3y = 0$ and $2x^2 + 2y^2 + 4x - 7y - 25 = 0$ and the point $(1, 1)$ is

The point at which the circles $x^2+y^2-4x-4y+7=0$ and $x^2+y^2-12x-10y+45=0$ touch each other is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module