Find the equation of the circle passing through the points $(4, 1)$ and $(6, 5)$ and whose centre is on the line $4x + y = 16$.

  • A
    $x^{2} + y^{2} - 6x - 8y + 15 = 0$
  • B
    $x^{2} + y^{2} - 6x - 8y + 10 = 0$
  • C
    $x^{2} + y^{2} - 6x - 8y + 20 = 0$
  • D
    $x^{2} + y^{2} - 6x - 8y + 25 = 0$

Explore More

Similar Questions

The equation of the circle touching $x = 0$,$y = 0$ and $x = 4$ is

Let a circle passing through $(2,0)$ have its centre at the point $(h, k)$. Let $(x_c, y_c)$ be the point of intersection of the lines $3x + 5y = 1$ and $(2+c)x + 5c^2y = 1$. If $h = \lim_{c \to 1} x_c$ and $k = \lim_{c \to 1} y_c$,then the equation of the circle is:

The abscissae of the two points $A$ and $B$ are the roots of the equation $x^2+2ax-b^2=0$ and their ordinates are roots of the equation $y^2+2py-q^2=0$. Then the equation of the circle with $AB$ as diameter is given by

$A$ circle passes through the points $(2, 3)$ and $(4, 5)$. If its centre lies on the line $y - 4x + 3 = 0$,then its radius is equal to

The equations of the two circles which touch the $Y$-axis at $(0,3)$ and make an intercept of $8$ units on the $X$-axis are

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