Question

A circle has its centre at the point
C[-2.3) and passes through the point PC-3,8). Find
an equation for this circle
A. (+ 2)+ (1 + 3)2 = 26
B. (2 + 2)2 + (y - 3)2 – 26
C. (2) + (21 - 3)2 = 26
D. (41 - 2)2 + (y - 3)? - 26​

Answer 1

Answer:

B

Step-by-step explanation:

the general equation of a circle with its centre (h,k) and radius r is given by

(x-a)²+(y-b)²=r²

given, the centre is (-2,3) and the circle also passes through (-3,8).

we know that radius is the distance between centre of circle and a point through which the circle passes.

Therefore,

$r = \sqrt{ {( (- 2) - ( - 3))}^{2} + {(3 - 8)}^{2} }$

$r = \sqrt{ {( - 2 + 3)}^{2} + {(3 - 8)}^{2} }$

$r = \sqrt{ {(1)}^{2} + {( - 5)}^{2} }$

$r = \sqrt{ 1 + 25 }$

$r = \sqrt{26}$

now the equation of circle is

${(x - ( - 2))}^{2} + {(y - 3)}^{2} = { (\sqrt{26}) }^{2}$

${(x + 2)}^{2} + {(y - 3)}^{2} = (26)$

therefore answer is option b

Please mark it as brainliest.

Thanks