Find the equation of the circumcircle of the | vedclass

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

Question

(A) The lines are $x = 0$ (y-axis),$y = 0$ (x-axis),and $\frac{x}{a} - \frac{y}{b} = 1$.The vertices of the triangle are the intersection points of th

Vedclass · Accepted Answer

$x^2 + y^2 - ax + by = 0$

Vedclass · Answer

(A) The lines are $x = 0$ (y-axis),$y = 0$ (x-axis),and $\frac{x}{a} - \frac{y}{b} = 1$.The vertices of the triangle are the intersection points of these lines:$1$. Intersection of $x=0$ and $y=0$ is $O(0, 0)$.$2$. Intersection of $x=0$ and $\frac{x}{a} - \frac{y}{b} = 1$ is $(0, -b)$.$3$. Intersection of $y=0$ and $\frac{x}{a} - \frac{y}{b} = 1$ is $(a, 0)$.Since the triangle is a right-angled triangle with vertices at $(0,0)$,$(a,0)$,and $(0,-b)$,the hypotenuse is the line segment connecting $(a,0)$ and $(0,-b)$.The circumcircle of a right-angled triangle has the hypotenuse as its diameter.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.Substituting $(a, 0)$ and $(0, -b)$:$(x - a)(x - 0) + (y - 0)(y - (-b)) = 0$$x(x - a) + y(y + b) = 0$$x^2 - ax + y^2 + by = 0$$x^2 + y^2 - ax + by = 0$.

Find the equation of the circumcircle of the triangle formed by the lines $x = 0$,$y = 0$,and $\frac{x}{a} - \frac{y}{b} = 1$.

Similar Questions

$A$ circle passes through the point $(-1, 0)$ and touches the $y$-axis at $(0, 2)$. Through which other point does this circle pass?

Find the equation of the circle passing through the origin and whose center is the intersection point of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$.

If the circle $x^{2}+y^{2}-2gx+6y-19c=0$,where $g, c \in R$,passes through the point $(6,1)$ and its centre lies on the line $x-2cy=8$,then the length of the intercept made by the circle on the $x$-axis is:

The radius of the circle $(x - 5)(x - 1) + (y - 7)(y - 4) = 0$ is

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154$ square units. The equation of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module