Let alpha, beta be the roots of x^2+5x+6=0 an | vedclass

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

Question

(C) The roots of $x^2+5x+6=0$ are $\alpha, \beta$. By Vieta's formulas,$\alpha+\beta = -5$ and $\alpha\beta = 6$.The roots of $y^2+6y+7=0$ are $\gamma

Vedclass · Accepted Answer

$x^2+y^2+5x+6y+13=0$

Vedclass · Answer

(C) The roots of $x^2+5x+6=0$ are $\alpha, \beta$. By Vieta's formulas,$\alpha+\beta = -5$ and $\alpha\beta = 6$.The roots of $y^2+6y+7=0$ are $\gamma, \delta$. By Vieta's formulas,$\gamma+\delta = -6$ and $\gamma\delta = 7$.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Here,the endpoints are $(\alpha, \gamma)$ and $(\beta, \delta)$.Thus,the equation is $(x-\alpha)(x-\beta) + (y-\gamma)(y-\delta) = 0$.Expanding this,we get $x^2 - (\alpha+\beta)x + \alpha\beta + y^2 - (\gamma+\delta)y + \gamma\delta = 0$.Substituting the values: $x^2 - (-5)x + 6 + y^2 - (-6)y + 7 = 0$.This simplifies to $x^2 + y^2 + 5x + 6y + 13 = 0$.

Let $\alpha, \beta$ be the roots of $x^2+5x+6=0$ and $\gamma, \delta$ be the roots of $y^2+6y+7=0$. Then the equation of the circle with $(\alpha, \gamma)$ and $(\beta, \delta)$ as the extremities of a diameter is

Similar Questions

The centre of the circle $r^2-4r(\cos \theta+\sin \theta)-4=0$ in Cartesian coordinates is

The equation $x^2 + y^2 = 0$ denotes

The equation of the circle,concentric with the circle $2x^2+2y^2-6x+8y+1=0$ and having double its area,is

The radius of the circle $r = \sqrt{3} \sin \theta + \cos \theta$ is

The number of integral values of $K$ for which $x^2+y^2+kx+(1-k)y+5=0$ represents a circle whose radius cannot exceed $5$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module