If the lines 2x + 3y + 1 = 0 and 3x - y - 4 = | vedclass

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

Question

(D) The center of the circle is the intersection point of its diameters. Solving the system of equations:$2x + 3y = -1$ $(i)$$3x - y = 4$ (ii)Multiply

Vedclass · Accepted Answer

$x^2 + y^2 - 2x + 2y - 23 = 0$

Vedclass · Answer

(D) The center of the circle is the intersection point of its diameters. Solving the system of equations:$2x + 3y = -1$ $(i)$$3x - y = 4$ (ii)Multiplying (ii) by $3$: $9x - 3y = 12$ (iii)Adding $(i)$ and (iii): $11x = 11 \Rightarrow x = 1$.Substituting $x = 1$ into (ii): $3(1) - y = 4 \Rightarrow y = -1$.Thus,the center $(h, k)$ is $(1, -1)$.Given the circumference $2\pi r = 10\pi$,we find the radius $r = 5$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.$(x - 1)^2 + (y + 1)^2 = 5^2$$x^2 - 2x + 1 + y^2 + 2y + 1 = 25$$x^2 + y^2 - 2x + 2y - 23 = 0$.

If the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ lie along diameters of a circle of circumference $10\pi$,then the equation of the circle is

Similar Questions

If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154 \text{ sq units}$,then the equation of the circle is (Take $\pi = \frac{22}{7}$)

$A$ circle passes through the centre of another circle $x^2+y^2-3x-4y-1=0$ and its centre is $(5,2)$. Then the equation of this circle is.........

If $x = \frac{2at}{1+t^2}$ and $y = \frac{a(1-t^2)}{1+t^2}$,where $t$ is a parameter,then $a$ represents:

The radius of a circle which touches the $y$-axis at $(0,3)$ and cuts an intercept of $8$ units with the $x$-axis is

The equation of the circle passing through the points $(0, 0)$,$(0, b)$,and $(a, b)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module