Question

A triangle has two of its sides along the axes, it's third side touches the circle x²+y²-4x-4y+4=0 locus of circumcentre of triangle is​

Answer 1

Answer:

Step-by-step explanation:

x²  + y² - 4x - 4y + 4 = 0

=> (x - 2)²  + (y - 2)² - 2² = 0

=> (x - 2)²  + (y - 2)² = 2²

hence center = 2 , 2  

and radius = 2

.Let third side be y = mx + c such that A is (p,0) and B is (0,q) and the line AB touches the given circle.

putting 0 , q    c = q => y = mx + q

putting p , 0   0 = mp + q   => m = -q/p

y = (-q/p)x  + q

dividing by q

y/q = -x/p + 1

=> x/p + y/q = 1

let say it touches circle at (h.k)

h/p + k/q = 1 => qh + pk = pq

(k - 2)/(h - 2) =  - 1/(-q/p)

=> (k - 2)/(h - 2) = p/q

Also (k-2)² + (h-2)² = 2²

(k - 2)/(h - 2) = p/q

=> qk - 2q = ph - 2p

=> qk - ph = 2(q - p)

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