CircleHard

Question

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

Options

A.x2 + y2 = 9a2
B.x2 + y2 = 16a2
C.x2 + y2 = 4a2
D.x2 + y2 = a2

Solution

           
Let ABC be an equilateral triangle, whose median is AD.
Given AD = 3a
In ᐃABD, AB2 = AD2 + BD2 ;
⇒ x2 = 9a2 + (x2/4) whereAB = BC = AC = x. x2 = 9a2 ⇒ x2 = 12a2
In ᐃOBD, OB2 = OD2 + BD2
⇒ r2 = (3a - r)2 + ⇒ r2 = 9a2 - 6ar + r2 + 3a2 ; ⇒ 6ar = 12a2 ⇒ r = 2a
So equation of circle is x2 + y2 = 4a2

Create a free account to view solution

View Solution Free
Topic: Circle·Practice all Circle questions

More Circle Questions

The length of the common chord of circle x2 + y2 − 6x − 16 = 0 and x2 + y2 − 8y − 9 = 0 is-...The equation of the lines joining the vertex of the parabola y2 = 6x to the points on it whose abscissa is 24, is -...If the distance between a tangent to the parabola y2 = 4 x and a parallel normal to the same parabola is 2√2, then...Pair of tangents are drawn from every point on the line 3x + 4y = 12 on the circle x2 + y2 = 4. Their variable chord of ...The common chord of x2+ y2 − 4x − 4y = 0 and x2 + y2 = 16 subtends at the origin an angle equal to-...