If P and Q are the points of intersection of | vedclass

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

Question

(D) The equation of the common chord (radical axis) of the two circles $S_1: x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ and $S_2: x^2 + y^2 + 2x + 2y - p^2 = 0

Vedclass · Accepted Answer

For no value of $p$

Vedclass · Answer

(D) The equation of the common chord (radical axis) of the two circles $S_1: x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ and $S_2: x^2 + y^2 + 2x + 2y - p^2 = 0$ is given by $S_1 - S_2 = 0$.$(x^2 + y^2 + 3x + 7y + 2p - 5) - (x^2 + y^2 + 2x + 2y - p^2) = 0$$x + 5y + 2p - 5 + p^2 = 0 \ldots (i)$$A$ circle passing through the intersection points $P$ and $Q$ of two circles $S_1$ and $S_2$ is given by the family of circles $S_1 + \lambda(S_1 - S_2) = 0$.For the circle to pass through the point $(1, 1)$,the point $(1, 1)$ must not lie on the common chord $PQ$ (unless the circles are identical).Substituting $(1, 1)$ into the equation of the common chord:$1 + 5(1) + 2p - 5 + p^2 = 0$$p^2 + 2p + 1 = 0$$(p + 1)^2 = 0 \Rightarrow p = -1$.If $p = -1$,the point $(1, 1)$ lies on the common chord,meaning the points $P, Q,$ and $(1, 1)$ are collinear,and no circle can pass through three collinear points. Thus,for any value of $p 
eq -1$,a unique circle exists.

If $P$ and $Q$ are the points of intersection of the circles $x^2 + y^2 + 3x + 7y + 2p - 5 = 0$ and $x^2 + y^2 + 2x + 2y - p^2 = 0$,then for what value of $p$ does a circle pass through $P, Q$ and $(1, 1)$?

Similar Questions

The ratio of the areas of the greatest and the smallest circles touching $(x \pm 1)^2 + (y \pm 1)^2 = 1$ is

If the circles $x^2 + y^2 + 2ax + c = 0$ and $x^2 + y^2 + 2by + c = 0$ touch each other,then:

The limiting points of the co-axial system containing the two circles $x^2+y^2+2x-2y+2=0$ and $25(x^2+y^2)-10x-80y+65=0$ are

The circle ${x^2 + y^2 + 2gx + 2fy + c = 0}$ bisects the circumference of the circle ${x^2 + y^2 + 2g'x + 2f'y + c' = 0}$,if

Let $S \equiv x^2+y^2-6x-6y+4=0$ and $S^{\prime} \equiv x^2+y^2-2x-4y+3=0$ be two circles. The centre of a circle of radius $\sqrt{14}$ and having the same radical axis as $S=0$ and $S^{\prime}=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module