The equation of a circle passing through the | vedclass

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

Question

(B) The centre of the circle is $(h, k) = (2, 2)$.Since the circle passes through the point $(4, 5)$,the radius $r$ is the distance between the centre

Vedclass · Accepted Answer

${x^2} + {y^2} - 4x - 4y - 5 = 0$

Vedclass · Answer

(B) The centre of the circle is $(h, k) = (2, 2)$.Since the circle passes through the point $(4, 5)$,the radius $r$ is the distance between the centre and the point:$r = \sqrt{(4 - 2)^2 + (5 - 2)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.The standard equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values,we get:$(x - 2)^2 + (y - 2)^2 = (\sqrt{13})^2$Expanding the terms:$(x^2 - 4x + 4) + (y^2 - 4y + 4) = 13$Simplifying the equation:$x^2 + y^2 - 4x - 4y + 8 = 13$$x^2 + y^2 - 4x - 4y - 5 = 0$.

The equation of a circle passing through the point $(4, 5)$ and having the centre at $(2, 2)$ is

Similar Questions

The equation of the circle in the first quadrant which touches each axis at a distance $5$ from the origin is

The abscissae of two points $P$ and $Q$ are the roots of the equation $2x^2 + 4x - 7 = 0$ and their ordinates are the roots of the equation $3x^2 - 12x - 1 = 0$. Then the centre of the circle with $PQ$ as a diameter is

The equation of the circle with center $(-a, -b)$ and radius $\sqrt{a^{2}-b^{2}}$ is:

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

What is the length of the diameter of the circle $x^2 + y^2 - 4x - 6y + 4 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module