Question

fined the coodintas of centre
and radius of the circle
x² + y² - 6x - 4y = 25​

Answer 1

Equation of circle with center (h,k) and radius r :-

$\boxed{ \rm{(x - h)}^{2} + {(y - k)}^{2} = {r}^{2} }....(1)$

Now we have to find co-ordinates of centre

and radius of the circle :-

$\rm \longrightarrow {x}^{2} + {y}^{2} - 6x - 4y = 25$

To find the co-ordinates of centre and radius of the given circle we should transform the given equation in the form of equation (1).

$\rm \longrightarrow {x}^{2} - 6x + {y}^{2} - 4y = 25 \\ \\ \\ \rm \longrightarrow {x}^{2} - 2(3)x + {(3)}^{2} - {(3)}^{2} + {y}^{2} - 2(2)y + {(2)}^{2} - {(2)}^{2} = 25\\ \\ \\ \rm \longrightarrow {(x - 3)}^{2} - {(3)}^{2} + {(y - 2)}^{2} - {(2)}^{2} = 25 \\ \\ \\ \rm \longrightarrow {(x - 3)}^{2} - 9 + {(y - 2)}^{2} - 4 = 25\\ \\ \\ \rm \longrightarrow {(x - 3)}^{2} + {(y - 2)}^{2} - 13 = 25\\ \\ \\ \rm \longrightarrow {(x - 3)}^{2} + {(y - 2)}^{2} - 13 = 25\\ \\ \\ \rm \longrightarrow {(x - 3)}^{2} + {(y - 2)}^{2} = 25 + 13\\ \\ \\ \rm \longrightarrow {(x - 3)}^{2} + {(y - 2)}^{2} = 38\\ \\ \\ \rm \longrightarrow {(x - 3)}^{2} + {(y - 2)}^{2} = {( \sqrt{38} )}^{2}$

➡️Co-ordinates of centre :-. (3,2)

➡️ Radius = √(38) units