Given:
Equation of circle is:
[tex](x+5)^2+(y-9)^2=8^2[/tex]
Find-:
Which point lies in the circle
[tex]\begin{gathered} A.(0,8) \\ \\ B.(13,-9) \\ \\ C.(-5,1) \\ \\ D.(3,17) \end{gathered}[/tex]
Explanation-:
If the points lie on a circle then it's true for the equation then,
(A)
Point is (0,8)
[tex](x,y)=(0,8)[/tex]
Check for the equation is:
[tex]\begin{gathered} (x+5)^2+(y-9)^2=8^2 \\ \\ (0+5)^2+(8-9)^2=8^2 \\ \\ 25+1=64 \\ \\ 26\ne64 \end{gathered}[/tex]
So it is a not a point in circle
(b)
Point is (13,-9)
[tex]\begin{gathered} (x+5)^2+(y-9)^2=8^2 \\ \\ (13+5)^2+(-9-9)^2=8^2 \\ \\ 18^2+(-18)^2\ne8^2 \end{gathered}[/tex]
So it is a not a point in circle
(c)
Point (-5,1)
[tex]\begin{gathered} (x+5)^2+(y-9)^2=8^2 \\ \\ (-5+5)^2+(1-9)^2=8^2 \\ \\ 0^2+(-8)^2=8^2 \\ \\ 64=64 \end{gathered}[/tex]
So, the point (-5,1) lies on circle.
