Question

1. which of the following points lies on the circle with equation (x - 4)^2+(y + 2)^2 = 36? a. (4,4) b. (0,2) c. (4,-2) d. (4,8) 6. which type of triangle is the circum - center located outside of? a. isosceles triangle b. obtuse triangle c. acute triangle d. right triangle 7. if the equation of a circle is x^2 + y^2 = 144, what is the radius of the circle? a. 72 b. 144 c. 12 d. 36

Explanation:

Step1: Recall circle - point relationship

For a circle with equation \((x - a)^2+(y - b)^2=r^2\), a point \((x_0,y_0)\) lies on the circle if \((x_0 - a)^2+(y_0 - b)^2=r^2\). For the circle \((x - 4)^2+(y + 2)^2 = 36\), we check each point:

• For point \((4,4)\): \((4 - 4)^2+(4+2)^2=0 + 36=36\).
• For point \((0,2)\): \((0 - 4)^2+(2 + 2)^2=16 + 16=32

eq36\).

• For point \((4,-3)\): \((4 - 4)^2+(-3 + 2)^2=0+1 = 1

eq36\).

• For point \((4,8)\): \((4 - 4)^2+(8 + 2)^2=0 + 100=100

eq36\).

Step2: Recall circum - center property

The circum - center of an obtuse - angled triangle is located outside the triangle. The circum - center of an acute - angled triangle is inside the triangle, and the circum - center of a right - angled triangle is on the hypotenuse. An isosceles triangle can be acute, right or obtuse, so its circum - center's position is not always outside.

Step3: Recall circle equation and radius

The standard form of a circle equation is \(x^{2}+y^{2}=r^{2}\), where \(r\) is the radius. Given \(x^{2}+y^{2}=144\), then \(r^{2}=144\), and \(r = 12\) (since \(r>0\)).

Answer:

1. A. \((4,4)\)
2. B. Obtuse triangle
3. C. 12