The equation of the circumcircle of the quadr | vedclass

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

Question

(B) The lines $x = a, x = 2a, y = -a, y = a$ form a rectangle with vertices at $(a, -a), (2a, -a), (2a, a),$ and $(a, a)$.Since the quadrilateral is a

Vedclass · Accepted Answer

$x^2 + y^2 - 3ax + a^2 = 0$

Vedclass · Answer

(B) The lines $x = a, x = 2a, y = -a, y = a$ form a rectangle with vertices at $(a, -a), (2a, -a), (2a, a),$ and $(a, a)$.Since the quadrilateral is a rectangle,its circumcircle has the diagonal as its diameter.Taking the diagonal connecting $(a, -a)$ and $(2a, a)$,the equation of the circle is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting the coordinates: $(x - a)(x - 2a) + (y - (-a))(y - a) = 0$.Expanding this: $(x^2 - 3ax + 2a^2) + (y + a)(y - a) = 0$.$(x^2 - 3ax + 2a^2) + (y^2 - a^2) = 0$.$x^2 + y^2 - 3ax + a^2 = 0$.

The equation of the circumcircle of the quadrilateral formed by the lines $x = a, x = 2a, y = -a, y = a$ is:

Similar Questions

The equation of the circle that touches the $Y-$axis at a distance of $4$ units from the origin and cuts off an intercept of $6$ units on the $X-$axis is

If a circle with center $(0, 0)$ touches the line $5x + 12y = 1$,then its equation is:

The equation of the circle with centre $(2, 1)$ and touching the line $3x + 4y = 5$ is

The equation of the circle passing through the point $(1, 1)$ and having two diameters along the pair of lines $x^2 - y^2 - 2x + 4y - 3 = 0$ is

The centre and radius of the circle $x^2+y^2+8x+10y-8=0$ respectively are

Vedclass Test Series

Exam Paper Generator

Online Exam Module