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

The lines $2x - 3y = 5$ and $3x - 4y = 7$ are the diameters of a circle of area $154$ square units. The equation of the circle is

$A$ circle passing through $(0,0)$, $(2,6)$, and $(6,2)$ cuts the $x$-axis at the point $P \neq (0,0)$. Then, the length of $OP$, where $O$ is the origin, is

The four distinct points $(0,0), (2,0), (0,-2)$ and $(k,-2)$ are concyclic,if $k$ is equal to

The equation of the circle that touches the $Y-$axis at a distance of $4$ units from the origin and cuts off an intercept of $6$ units on the $X-$axis is

The equation of the circle which passes through the points $(3, -2)$ and $(-2, 0)$ and whose centre lies on the line $2x - y = 3$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module