Question

The circle below has center P.The point (x, y) is on the circle as shown.12-(a) Find the following.1110-unitsRadius: 0Center: 0987Value of a:(Choose one)(x,y)351Value of b:(Choose one)4a32(b) Use the Pythagorean Theorem to write an equationrelating the side lengths of the right triangle. Writeyour answer in terms of x and y (with no otherletters)+

Answer 1

The radius of the circle is 3 units, the value of a is
`√(5)` units and the value of b is 2 units. The Pythagoras theorem to the equations relating the side lengths of the right triangle in terms of x and y is
`x^2+y^2=9`.

The radius of a circle is any of the line segments from its center to its perimeter. As, we can clearly see in the image, radius= 3 units.

The Point (P) is the center of the circle. If we draw a straight light line toward the x-axis and toward the y-axis, we can see that center P=(9,4)

The value of b is for 4 units to 6 units towards the y-axis direction. So, the value of b is 2 units.

The value of a can be found by using Pythagoras theorem:

`3^2=a^2+2^2`

`a^2=9-4`

`a=√(5)`

So, the value of a is
`√(5)`.

b) The Pythagoras theorem to the equations relating the side lengths of the right triangle in terms of x and y is:

`x^2+2^2=3^2` or

`x^2+y^2=3^2` (as we want side lengths to be x and y)

Answer 2

Given:

Center of the circle = P

Let's determine the following:

a) Radius.

Here, the radius of the circle is the hypotenuse of the triangle.

Therefore, the radius of the circle is 3 units

b) Center:

To find the point at the center of the circle, let's locate the point P on the graph.

On the graph, the point P is at (x, y) ==> (9, 4)

Therefore, the center (h, k) is (9, 4)

c) Value of a:

To find the value of a, let's first find the value of b.

Value of b = 6 - 4 = 2

Apply Pythagorean Theorem to find the value of a:

`c^2=a^2+b^2`

Where:

c is the hypotenuse = 3

b = 2

Thus, we have:

`\begin{gathered} 3^2=a^2+2^2 \\ \\ 9=a^2+4 \\ \\ \text{Subtract 4 from both sides:} \\ 9-4=a^2+4-4 \\ \\ 5=a^2 \\ \\ \text{Take the square root of both sides:} \\ \sqrt[]{5}=\sqrt[]{a^2} \\ \\ 2.2=a \\ \\ a=2.2 \end{gathered}`

Therefore, the value of a is 2.2 units

d) Value of b.

The value of b is 2 units

ANSWERS:

• Radius: , 3 units

,

• Center: , (9, 4)

,

• Value of a = , 2.2 units

,

• Value of b = , 2 units