Let P(x_1, y_1) and Q(x_2, y_2) be two points | vedclass

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

Question

(A) The abscissae $x_1$ and $x_2$ are the roots of the equation $x^2 + 2x - 3 = 0$.Using the sum of roots formula,$x_1 + x_2 = -\frac{b}{a} = -2$.The

Vedclass · Accepted Answer

$(-1, -2)$

Vedclass · Answer

(A) The abscissae $x_1$ and $x_2$ are the roots of the equation $x^2 + 2x - 3 = 0$.Using the sum of roots formula,$x_1 + x_2 = -\frac{b}{a} = -2$.The $x$-coordinate of the centre is $\frac{x_1 + x_2}{2} = \frac{-2}{2} = -1$.The ordinates $y_1$ and $y_2$ are the roots of the equation $y^2 + 4y - 12 = 0$.Using the sum of roots formula,$y_1 + y_2 = -\frac{b}{a} = -4$.The $y$-coordinate of the centre is $\frac{y_1 + y_2}{2} = \frac{-4}{2} = -2$.Thus,the centre of the circle with $PQ$ as diameter is $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) = (-1, -2)$.

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two points such that their abscissae $x_1$ and $x_2$ are the roots of the equation $x^2 + 2x - 3 = 0$,while the ordinates $y_1$ and $y_2$ are the roots of the equation $y^2 + 4y - 12 = 0$. The centre of the circle with $PQ$ as diameter is

Similar Questions

The equation of the circle passing through $(0,0)$ and which makes intercepts $a$ and $b$ on the coordinate axes is

Find the equation of the circle with center at $(0,0)$ and radius $r$.

The centre of the circle passing through the point $(0,1)$ and touching the curve $y=x^2$ at $(2,4)$ is

The equation of the circle touching the lines $|x-2|+|y-3|=4$ is

The equation of the circle with centre $(2, -3)$ and touching the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module