Given:
Equation of circle is:
![(x+5)^2+(y-9)^2=8^2](https://img.qammunity.org/2023/formulas/mathematics/college/d7g6em30z7wmnxi1y2pxsqrxjhy37dplgg.png)
Find-:
Which point lies in the circle
![\begin{gathered} A.(0,8) \\ \\ B.(13,-9) \\ \\ C.(-5,1) \\ \\ D.(3,17) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ujxiiy60k8wn49uyq1v709der51ztbw18l.png)
Explanation-:
If the points lie on a circle then it's true for the equation then,
(A)
Point is (0,8)
![(x,y)=(0,8)](https://img.qammunity.org/2023/formulas/mathematics/college/skfxcvln5t7kulenob5hmgjax7kw6ivpxi.png)
Check for the equation is:
![\begin{gathered} (x+5)^2+(y-9)^2=8^2 \\ \\ (0+5)^2+(8-9)^2=8^2 \\ \\ 25+1=64 \\ \\ 26\\e64 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8lpxizzazg4yr051q2tn6acsweiuzsqxsp.png)
So it is a not a point in circle
(b)
Point is (13,-9)
![\begin{gathered} (x+5)^2+(y-9)^2=8^2 \\ \\ (13+5)^2+(-9-9)^2=8^2 \\ \\ 18^2+(-18)^2\\e8^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mzz50bo3i0hqp3yhtmnuvyop4zda0ce9w9.png)
So it is a not a point in circle
(c)
Point (-5,1)
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/z4ulegxs2ig73osocr6crdv2i534velknw.png)
So, the point (-5,1) lies on circle.