The equation of the circle with centre (2, -3 | vedclass

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

Question

(C) The centre of the circle is $(h, k) = (2, -3)$.Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coor

Vedclass · Accepted Answer

$x^2+y^2-4x+6y+4=0$

Vedclass · Answer

(C) The centre of the circle is $(h, k) = (2, -3)$.Since the circle touches the $X$-axis,the radius $r$ is equal to the absolute value of the $y$-coordinate of the centre.$r = |k| = |-3| = 3$.The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get $(x - 2)^2 + (y - (-3))^2 = 3^2$.$(x - 2)^2 + (y + 3)^2 = 9$.Expanding the terms: $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 9$.$x^2 + y^2 - 4x + 6y + 4 + 9 = 9$.$x^2 + y^2 - 4x + 6y + 4 = 0$.

The equation of the circle with centre $(2, -3)$ and touching the $X$-axis is

Similar Questions

The radius of the circle $r=12 \cos \theta+5 \sin \theta$ 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

If a circle passes through points $(4,0)$ and $(0,2)$ and its centre lies on the $Y$-axis. If the radius of the circle is $r$,then the value of $r^2-r+1$ is

Find the centre and the radius of the circle $x^{2}+y^{2}+8x+10y-8=0$.

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 Test Series

Exam Paper Generator

Online Exam Module