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

  • A
    $x^2 + y^2 + ax + by = 0$
  • B
    $x^2 + y^2 - ax + by = 0$
  • C
    $x^2 + y^2 - ax - by = 0$
  • D
    $x^2 + y^2 + ax - by = 0$

Explore More

Similar Questions

The equation of the circle whose centre lies on the lines $3x - y - 4 = 0$ and $x + 3y + 2 = 0$ and has an area of $154$ square units is:

$A$ line meets the coordinate axes at points $A$ and $B$. $A$ circle is circumscribed about the triangle $OAB$. If $m$ and $n$ are the distances of the tangents to the circle at points $A$ and $B$ respectively from the origin,then the diameter of the circle is

Difficult
View Solution

The equation of the circle whose centre lies on the line $x-4y=1$ and which passes through the points $(3,7)$ and $(5,5)$ 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

An infinite number of tangents can be drawn from $(1, 2)$ to the circle ${x^2} + {y^2} - 2x - 4y + \lambda = 0$. Then,$\lambda = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo