Question

Find equation of circle
passing through (2,7) radius = 13 , abscissa
of the centre =-3.​

Answer 1

Answer:

(x + 3)² + (y - 19)² = 13² and (x + 3)² + (y + 5)² = 13²

Step-by-step explanation:

Equation of a circle is (x - a)² + (y - b)² = r² , centered at (a,b) and with radius "r".

A( $x_{1}$ , $y_{1}$ )

B( $x_{2}$ , $y_{2}$ )

AB = $\sqrt{(x_{2} -x_{1})^2 +(y_{2} -y_{1} )^2}$

~~~~~~~~~~~

$\sqrt{(2+3)^2 +(7 - y)^2}$ = 13

25 + (7 - y)² = 169

(7 - y)² = 144

7 - y = ± 12

$y_{12}$ = 7 ± 12

$y_{1}$ = 19

$y_{2}$ = - 5

There are two circles with given output.

(x + 3)² + (y - 19)² = 13²

(x + 3)² + (y + 5)² = 13²