The standard equations of circle A are;
- [tex]\left(x - \dfrac{2535}{209} \right)^2 + y^2 = \left( \dfrac{1690}{209} \right)^2[/tex]
Method for deriving the above standard equations of a circle;
Known;
The general for of the equation of a circle is (x - h)² + (y - k)² = r²
The radius of circle C = 4·r
Where;
r = The radius of circle O
The point on circle C = (63, 16)
Solution:
The above values and equation gives;
(63 - h)² + (16 - k)² = (4·r)²
k = The y-coordinate of the center of circle C which is on the x-axis
Therefore;
k = 0 = The y-coordinate of all circles in the diagram
h = r + 4·r + 6·r + 4·r = 15·r
(63 - 15·r)² + (16 - 0)² = (4·r)²
Which gives;
209·r² - 1,890·r + 4225 = 0
(r - 5)·(209·r - 845) = 0
r = 5 or r = [tex]\dfrac{845}{209}[/tex]
The radius of circle A, [tex]r_A[/tex] = 2·r
Therefore;
[tex]r_A[/tex] = 2 × 5 = 10
The x-coordinate of the center of circle A = 3·r
Therefore;
The center of circle A = (3×5, 0) = (15, 0)
- The standard equation of circle A is (x - 15)² + y² = 10²
[tex]When \ r = \mathbf{\dfrac{845}{209}}[/tex]
[tex]r_A = 2 \times \dfrac{845}{209} = \dfrac{1690}{209}[/tex]
[tex]Coordinates \ of \ A =\left(3 \times \dfrac{845}{209}, \ 0\right) = \left(\dfrac{2535}{209} , \ 0 \right)[/tex]
- [tex]The \ other\ standard \ equation \ of \ circle \ A \ is; \underline{\left(x - \dfrac{2535}{209}\right)^2 + y^2 = \left(\dfrac{1690}{209} \right)^2}[/tex]
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