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Sagot :
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:
[tex]c^2=a^2+b^2[/tex]Where:
c is the hypotenuse = 3
b = 2
Thus, we have:
[tex]\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}[/tex]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
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