Question

A triangle has two sides along the coordinate axes and the third side is a tangent to the circle

${x}^{2} + {y }^{2} - 2ax - 2ay + {a}^{2} = 0$
Find the locus of the centroid of the triangle .The answer is in the image . Please explain .​

Answer 1

Answer:

x4÷3÷4÷6÷9÷×

Step-by-step explanation:

x4÷3÷4÷6÷9÷×

Answer 2

Answer:

Step-by-step explanation:

x²  + y² - 2ax - 2ay + a² = 0

=> (x - a)²  + (y - a)² - a² = 0

=> (x - a)²  + (y - a)² = a²

hence center = a , a  

and radius = a

.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 - a)/(h - a) =  - 1/(-q/p)

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

Also (k-a)² + (h-a)² = a²

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

=> qk - qa = ph - pa

=> qk - ph = qa - pa