Question

. If an equilateral triangle is inscribed in the circle x² + y2 - 6x – 4y + 5 = 0, then its side ​

Answer 1

Answer:

if centre= (p,q) and radius =r

Then equation of the circle would be

(x-p)^2 + (y-q)^2 = r^2. ———-> (i)

Given eqn is

x^2 + y^2 -6x-4y+5=0

=> x^2 - 2*3*x +3^2 +y^2- 2*2*y +2^2 +5–3^2–2^2=0

=> (x-3)^2 + (y-2)^2= 8 —-> (ii)

From eqn (i) and ((ii)

r^2= 8 ———->(iii)

Now, if a be the side of equilateral triangle and that is inscribe in circle with radius r.

Then,

r^2 =(a^2)/3——-(iv)

From eqn (iii) and (iv)

a^2 = 3* r^2= 3* 8= 24

So length of triangle = a = √24 = 2√6.

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