$A$ circle has a radius of $3$ units and its centre lies on the line $y = x - 1$. If the circle passes through the point $(7, 3)$,what is its equation?

  • A
    ${x^2} + {y^2} - 8x - 6y + 16 = 0$
  • B
    ${x^2} + {y^2} + 8x + 6y + 16 = 0$
  • C
    ${x^2} + {y^2} - 8x - 6y - 16 = 0$
  • D
    None of these

Explore More

Similar Questions

The equation of the circle passing through the origin and cutting intercepts of length $3$ and $4$ units from the positive axes is:

The equation obtained by transforming $x^2+y^2-6x+10y-2=0$ to the parallel axes through $(3,-5)$ is

The radius of the circle $x^2 + y^2 + 4x + 6y + 13 = 0$ is

If the lines $2x - 3y = 5$ and $3x - 4y = 7$ are two diameters of a circle of radius $7$,then the equation of the circle is

Find the equation of a circle with radius $4$ that touches the $x$-axis at a distance of $-3$ from the origin.

Difficult
View Solution

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