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

If the equation of the circle passing through the points $(-1,0), (-1,1), (1,1)$ is $ax^2+ay^2+2gx+2fy-2=0$,then $a=$

Find the Cartesian equation of the circle whose parametric equations are $x = -7 + 4 \cos \theta$ and $y = 3 + 4 \sin \theta$.

The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ is

The number of circles touching the lines $x = 0$,$y = a$ and $y = b$ is

If the lines $x + y = 6$ and $x + 2y = 4$ are diameters of a circle whose diameter is $20$,then the equation of the circle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module