Which of the following is the equation of a c | vedclass

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

Question

(D) The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$,where the coefficients of $x^2$ and $y^2$ are equal and there is no $

Vedclass · Accepted Answer

$3(x^2 + y^2) + 5x + 1 = 0$

Vedclass · Answer

(D) The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$,where the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term.Option $A$ has different coefficients for $x^2$ and $y^2$ ($1$ and $2$).Option $B$ has different signs for $x^2$ and $y^2$.Option $C$ contains an $xy$ term.Option $D$ can be written as $3(x^2 + y^2) + 5x + 1 = 0$,which simplifies to $x^2 + y^2 + \frac{5}{3}x + \frac{1}{3} = 0$. This matches the form of a circle equation.

Which of the following is the equation of a circle?

Similar Questions

If the lines $l_1x + m_1y + n_1 = 0$ and $l_2x + m_2y + n_2 = 0$ cut the axes at concyclic points,then

The abscissae of two points $P$ and $Q$ are the roots of the equation $2x^2 + 4x - 7 = 0$ and their ordinates are the roots of the equation $3x^2 - 12x - 1 = 0$. Then the centre of the circle with $PQ$ as a diameter is

Circles are drawn through the point $(2,0)$ to cut intercepts of length $5$ units on the $X$-axis. If their centre lies in the first quadrant,then their equation is

The parametric equations of the circle $2x^2 + 2y^2 = 9$ are

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module