Respuesta :
Answer
Explanations:
1a) The standard form of the equation of a circle is expressed using the formula shown below;
[tex](x-a)^2+(y-b)^2=r^2[/tex]where:
(a, b) is the centre of the circle
"r" is the radius of the circle
The radius of the circle will be the distance from the center (3,-7) and the point located on the circle at (-4,1). This can be gotten using the distance formula as shown:
[tex]\begin{gathered} r=\sqrt[]{(x_2-x_1)^2_{}+(y_2-y_1)^2} \\ r=\sqrt[]{(-4-3)^2+(1-(-7))}^2 \\ r=\sqrt[]{(-7)^2+8^2} \\ r=\sqrt[]{49+64} \\ r=\sqrt[]{113} \end{gathered}[/tex]Substitute the radius and the centre (3, -7) into the formula for the equation of a circle as shown:
[tex]\begin{gathered} (x-3)^2+(y-(-7))^2=(\sqrt[]{113})^2 \\ (x-3)^2+(y+7)^2=113^{} \end{gathered}[/tex]This given the required equation of the circle
2) In order to check whether the point (7, 6) lies on the circle, we will simply substitute the coordinate point in the resulting equation as shown:
[tex](7-3)^2+(6+7)^2=4^2+13^2=185\neq113[/tex]Since the result is not equal to 113, this shows that the point (7,6) is NOT on the circle
