Circles, Ellipses, Conic Sections — Quiz 6

Questions in this quiz (8)

1. In the standard $$(x,y)$$ coordinate plane, a circle determined by $$x^2+4x+y^2-2y+1=0$$ is translated 3 units to the left and 2 units up. Which of the following gives an equation of the translated circle?

   • $$(x-3)^2+(y+1)^2=4$$
   • $$(x+1)^2+(y-3)^2=4$$
   • $$(x-1)^2+(y+3)^2=4$$
   • $$(x+3)^2+(y-1)^2=4$$

2. A circle in the standard $$(x,y)$$ coordinate plane is tangent to the $$x$$-axis at 5 and tangent to the $$y$$-axis at 5. Which of the following is an equation of the circle?

   • $$(x+5)^2+(y+5)^2=5$$
   • $$(x-5)^2+(y-5)^2=25$$
   • $$x^2+y^2=25$$
   • $$(x+5)^2+(y+5)^2=25$$

3. A circular garden is designed with its center at $$(7,-4)$$ in a coordinate plane. The garden is designed such that it touches the x-axis. What is the equation of the garden's boundary?

   • $$(x-7)^2+(y+4)^2=16$$
   • $$(x+7)^2+(y+4)^2=4$$
   • $$(x-7)^2+(y-4)^2=4$$
   • $$(x-7)^2+(y+4)^2=16$$

4. Point $$B$$ is outside a circle. The longest distance between the point and the circle is $$12$$ and the shortest is $$6$$. Find the radius of the circle.

   • 9
   • 12
   • 3
   • 6

5. Rotate the circle clockwise around its center by 45°:$$(x+2)^2+(y-5)^2=4^2$$Write the equation of the new circle.

   • $$(x-2)^2+(y-5)^2=16$$
   • $$(x+2)^2+(y-5)^2=16$$
   • $$(x-2)^2+(y+5)^2=16$$
   • $$(x+2)^2+(y+5)^2=16$$

6. Write the equation of a circle that passes through the three points: $$P(1,1), Q(1,5), R(5,1)$$

   • $$(x+3)^2+(y+3)^2=4$$
   • $$(x-5)^2+(y+5)^2=4$$
   • $$(x-3)^2+(y-3)^2=8$$
   • $$(x+5)^2+(y+5)^2=4$$

7. Write the new equation after the operation below on the following ellipse $$\frac{(x+2)^2}{3^2}+\frac{(y-4)^2}{6^2}=1$$Rotate it clockwise around its center 90°.

   • $$\frac{(x+4)^2}{6^2}+\frac{(y+2)^2}{3^2}=1$$
   • $$\frac{(x-4)^2}{6^2}+\frac{(y-2)^2}{3^2}=1$$
   • $$\frac{(x-4)^2}{6}+\frac{(y-2)^2}{3}=1$$
   • $$\frac{(x-4)^2}{6}+\frac{(y+2)^2}{3}=1$$

8. Write the new equation after the operation below on the following ellipse $$\frac{(x+2)^2}{5^2}+\frac{(y-1)^2}{3^2}=1$$Rotate it clockwise around the origin by 90°.

   • $$\frac{(2-y')^2}{25}+\frac{(x'-1)^2}{9}=1$$
   • $$\frac{(2+y')^2}{25}+\frac{(x'+1)^2}{9}=1$$
   • $$\frac{(2+y')^2}{25}+\frac{(x'-1)^2}{9}=1$$
   • $$\frac{(2-y')^2}{25}+\frac{(x'+1)^2}{9}=1$$