If the lines x+2y-5=0 and 2x-3y+4=0 lie along | vedclass

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

Question

(A) Given: The lines $x+2y-5=0$ and $2x-3y+4=0$ are diameters of the circle. Since the intersection of diameters is the center of the circle,we solve

Vedclass · Accepted Answer

$x^2+y^2-2x-4y-4=0$

Vedclass · Answer

(A) Given: The lines $x+2y-5=0$ and $2x-3y+4=0$ are diameters of the circle. Since the intersection of diameters is the center of the circle,we solve the system of equations: $x+2y=5$ $(i)$ $2x-3y=-4$ (ii) Multiplying $(i)$ by $2$,we get $2x+4y=10$ (iii). Subtracting (ii) from (iii): $(2x+4y)-(2x-3y) = 10-(-4)$ $\Rightarrow 7y=14$ $\Rightarrow y=2$. Substituting $y=2$ into $(i)$: $x+2(2)=5 \Rightarrow x=1$. Thus,the center $(h, k)$ is $(1, 2)$. The area of the circle is $9\pi$,so $\pi r^2 = 9\pi \Rightarrow r^2=9$. The equation of the circle is $(x-h)^2 + (y-k)^2 = r^2$. $(x-1)^2 + (y-2)^2 = 9 (x^2-2x+1) + (y^2-4y+4) = 9 x^2+y^2-2x-4y-4=0$.

If the lines $x+2y-5=0$ and $2x-3y+4=0$ lie along diameters of a circle of area $9\pi$,then the equation of the circle is

Similar Questions

If $(1, 1), (-2, 2), (2, -2)$ are $3$ points on a circle $S$,then the perpendicular distance from the centre of the circle $S$ to the line $3x - 4y + 1 = 0$ is

If the power of the point $(1, 6)$ with respect to the circle $x^2 + y^2 + 4x - 6y - a = 0$ is $-16$,then $a =$

If the vertices of a triangle are $(2, -2)$,$(-1, -1)$,and $(5, 2)$,then the equation of its circumcircle is

$A$ circle passes through the points $(1,2)$ and $(3,4)$. If its centre lies on the line $x-y+3=0$,then its radius is equal to

Find the equation of the circle with radius $5$ whose centre lies on the $x-$axis and passes through the point $(2, 3)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module