If the incenter of an equilateral triangle is | vedclass

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

Question

(A) For an equilateral triangle,the incenter and circumcenter coincide at the same point $(1, 1)$.The inradius $r$ is the perpendicular distance from

Vedclass · Accepted Answer

$x^2 + y^2 - 2x - 2y - 14 = 0$

Vedclass · Answer

(A) For an equilateral triangle,the incenter and circumcenter coincide at the same point $(1, 1)$.The inradius $r$ is the perpendicular distance from the incenter $(1, 1)$ to the side $3x + 4y + 3 = 0$:$r = \frac{|3(1) + 4(1) + 3|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 4 + 3|}{5} = \frac{10}{5} = 2$.In an equilateral triangle,the circumradius $R$ is twice the inradius $r$:$R = 2r = 2(2) = 4$.The equation of the circumcircle with center $(1, 1)$ and radius $R = 4$ is:$(x - 1)^2 + (y - 1)^2 = 4^2$$x^2 - 2x + 1 + y^2 - 2y + 1 = 16$$x^2 + y^2 - 2x - 2y - 14 = 0$.

If the incenter of an equilateral triangle is $(1, 1)$ and the equation of one of its sides is $3x + 4y + 3 = 0$,then the equation of the circumcircle of this triangle is

Similar Questions

Circles are drawn through the point $(2, 0)$ to cut an intercept of length $5$ units on the $x$-axis. If their centres lie in the first quadrant,then their equation is

$A$ circle touches the $y$-axis at the point $(0, 4)$ and cuts the $x$-axis in a chord of length $6$ units. The radius of the circle is

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

Find the equation of the circle passing through the origin and whose center is the intersection point of the lines $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$.

The equation of the circle passing through the points $(1, \sqrt{2})$,$(7, \sqrt{2})$,and $(1, 3)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module