The equation of the circle with centre (1, 2) | vedclass

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

Question

(B) The radius of the circle is the perpendicular distance from the centre $(1, 2)$ to the tangent line $x + y - 5 = 0$.$r = \frac{|1 + 2 - 5|}{\sqrt{

Vedclass · Accepted Answer

$x^2 + y^2 - 2x - 4y + 3 = 0$

Vedclass · Answer

(B) The radius of the circle is the perpendicular distance from the centre $(1, 2)$ to the tangent line $x + y - 5 = 0$.$r = \frac{|1 + 2 - 5|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$The equation of the circle with centre $(h, k) = (1, 2)$ and radius $r = \sqrt{2}$ is given by $(x - h)^2 + (y - k)^2 = r^2$.$(x - 1)^2 + (y - 2)^2 = (\sqrt{2})^2$$(x^2 - 2x + 1) + (y^2 - 4y + 4) = 2$$x^2 + y^2 - 2x - 4y + 5 = 2$$x^2 + y^2 - 2x - 4y + 3 = 0$Thus,the correct option is $B$.

The equation of the circle with centre $(1, 2)$ and tangent $x + y - 5 = 0$ is

Similar Questions

Find the equation of the circle with center at $(0,0)$ and radius $r$.

The equation of the circle concentric with the circle $x^2+y^2-6x-4y-12=0$ and touching the $Y$-axis is:

The circle touching the $y$-axis at a distance $4$ units from the origin and cutting off an intercept $6$ from the $x$-axis 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 the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ lie along diameters of a circle of circumference $10\pi$,then the equation of the circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module