Answer: The point that lies on the circle is (4) F(-2, 6).
Step-by-step explanation: We are to select the correct point that lies on a circle that that is centred at A(-3, 2) and passes through B(1, 3).
The standard equation of a circle with centre at (g, h) and radius 'r' units is given by
[tex](x-g)^2+(y-h)^2=r^2~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
The centre of the given circle is A(-3, 2), so we have
(g, h) = (-3, 2).
Also, the circle passes through the point B(1, 3).
Substituting these values in equation (i), we get
[tex](1-(-3))^2+(3-2)^2=r^2\\\\\Rightarrow 4^2+1=r^2\\\\\Rightarrow r^2=17\\\\\Rightarrow r=\sqrt{17}.[/tex]
Therefore, the equation of the circle with centre (-3, 2) and radius √17 units is given by
[tex](x-(-3))^2+(y-2)^2=(\sqrt{17})^2\\\\\Rightarrow (x+3)^2+(y-2)^2=17\\\\\Rightarrow x^2+6x+9+y^2-4y+4=17\\\\\Rightarrow x^2+y^2+6x-4y=4.[/tex]
Thus, the required equation of the circle is
[tex]x^2+y^2+6x-4y=4.[/tex]
Option (1) is C(-1, -2).
We have
[tex]x^2+y^2+6x-4y=(-1)^2+(-2)^2+6(-1)-4(-2)=1+4-6+8=7>4.[/tex]
So, the point C lies outside the circle.
Option (2) is D(-6, 3).
We have
[tex]x^2+y^2+6x-4y=(-6)^2+3^2+6(-6)-4(3)=36+9-36-12=-3<4.[/tex]
So, the point D lies inside the circle.
Option (3) is E(-3, -3).
We have
[tex]x^2+y^2+6x-4y=(-3)^2+(-3)^2+6(-3)-4(-3)=9+9-18+12=12>4.[/tex]
So, the point C lies outside the circle.
Option (4) is F(-2, 6).
We have
[tex]x^2+y^2+6x-4y=(-2)^2+6^2+6(-2)-4(6)=4+36-12-24=4.[/tex]
So, the point F lies on the circle.
Thus, the point that lies on the circle is F(-2, 6).
Option (4) is correct.
