Question

1. convert the following equation of a circle in general form to the standard form and then find the radius and the center.

$x^{2}-20x + y^{2}+4y + 55 = 0$
a. center: $(10,-2)$
radius: 7
b. center: $(10,-2)$
radius: 49
c. center: $(-2,10)$
radius: 7
d. center: $(-2,10)$
radius: 49

1. solve for the missing angle.

a. $80^{circ}$
b. $245^{circ}$
c. $65^{circ}$
d. $100^{circ}$

Explanation:

Step1: Group x and y terms

$$x^{2}-20x + y^{2}+4y=-55$$

Step2: Complete the square for x

For \(x^{2}-20x\), \((\frac{-20}{2})^{2}=100\). Add 100 to both sides.
$$x^{2}-20x + 100+y^{2}+4y=-55 + 100$$

Step3: Complete the square for y

For \(y^{2}+4y\), \((\frac{4}{2})^{2}=4\). Add 4 to both sides.
$$x^{2}-20x + 100+y^{2}+4y+4=-55 + 100+4$$

Step4: Write in standard form

$$(x - 10)^{2}+(y + 2)^{2}=49$$
The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. Here \(h = 10,k=-2,r=\sqrt{49}=7\)

Step1: Use the property of the angle formed by two tangents

The angle between a tangent and a radius at the point of tangency is \(90^{\circ}\). Let the center of the circle be \(O\). \(\angle OBC=\angle ODC = 90^{\circ}\)

Step2: Use the sum of angles in a quadrilateral

In quadrilateral \(ODCB\), the sum of interior angles is \(360^{\circ}\). Let \(\angle C\) be the unknown angle.
\(\angle C+90^{\circ}+115^{\circ}+90^{\circ}=360^{\circ}\)

Step3: Solve for \(\angle C\)

\(\angle C=360^{\circ}-(90^{\circ}+115^{\circ}+90^{\circ})=360^{\circ}-295^{\circ}=100^{\circ}\)

Answer:

A. Center: (10, -2) Radius: 7