Question

find the equation of the circle through the point (2,3) concentric with x2+y2-8x+4y+12​

Answer 1

Answer:

Step-by-step explanation:

Solution :-

Given equation of the circle is :-

$\sf x^2+y^2-8x+4y+12=0$

Comparing with general equation of the circle :- "ax^2 + by^2+2gx + 2fy + 2jxy + c = 0

$\sf \implies a=1,b=1,j=0,g=4,f=2,c=12$

(g , f) = (4 , 2) = (h , k)

Now passes through the point (2 , 3) :-

Radius(r) of the circle is :-

$r=\sqrt{(4-2)^2+(2-3)^2}$

$r=\sqrt{2^2+(-1)^2}$

$r=\sqrt{4+1}=\sqrt{5}$

Now equation of the circle :-

$(x-r)^2+(y-k)^2=r^2$

$(x-4)^2+(y-2)^2=(\sqrt{5})^2$

$(x-4)^2+(y-2)^2=5$