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Sagot :
Answer:
Radius of the circle would be [tex]10[/tex].
Center of the circle is at [tex](10,\, 10)[/tex].
Step-by-step explanation:
Let r denote the radius of the circle, and let [tex](a,\, b)[/tex] denote the center. A point [tex](x,\, y)[/tex] is on the circle if and only if the coordinates satisfy the equation:
[tex](x - a)^{2} + (y - b)^{2} = r^{2}[/tex].
(This equation is based on the distance formula between two points in a cartesian plane. In other words, the distance between [tex](x,\, y)[/tex] and the center of the circle should be equal to the radius of the circle.)
Obtain one equation about [tex]\text{$a$, $b$, and $r$}[/tex] for each point on this circle:
- [tex](10,\, 20)[/tex], where [tex]x = 10[/tex] and [tex]y = 20[/tex]: [tex](10 - a)^{2} + (20 - b)^{2} = r^{2}[/tex].
- [tex](16,\, 2)[/tex], where [tex]x = 16[/tex] and [tex]y = 2[/tex]: [tex](16 - a)^{2} + (2 - b)^{2} = r^{2}[/tex].
- [tex](2,\, 4)[/tex], where [tex]x = 2[/tex] and [tex]y = 4[/tex]: [tex](2 - a)^{2} + (4 - b)^{2} = r^{2}[/tex].
Rearrange the equations and solve for [tex]\text{$a$, $b$, and $r$}[/tex]. For example, subtract the first equation from the second to eliminate [tex]r[/tex]:
[tex](16 - a)^{2} + (2 - b)^{2} - (10 - a)^{2} - (20 - b)^{2} = 0[/tex].
Make use of the property [tex]A^{2} - B^{2} = (A - B)\, (A + B)[/tex] to simplify this equation:
[tex]\left((16 - a)^{2} - (10 - a)^{2}\right) + \left((2 - b)^{2} - (20 - b)^{2}\right) = 0[/tex].
[tex]\begin{aligned} & ((16 - a) - (10 - a))\, ((16 - a) + (10 - a)) \\ &+ ((2 - b) - (20 - b))\, ((2 - b) + (20 - b)) = 0\end{aligned}[/tex].
Simplify to obtain:
[tex](6)\, (26 - 2\, a) + (-18)\, (22 - 2\, b) = 0[/tex].
[tex](13 - a) + (-3)\, (11 - b) = 0[/tex].
[tex]-a + 3\, b = 20[/tex].
Similarly, subtract [tex](16 - a)^{2} + (2 - b)^{2} = r^{2}[/tex] from [tex](2 - a)^{2} + (4 - b)^{2} = r^{2}[/tex] to obtain:
[tex]\left((16 - a)^{2} - (2 - a)^{2}\right) + \left((2 - b)^{2} - (4 - b)^{2}\right) = 0[/tex].
[tex](14)\, (18 - 2\, a) + (-2)\, (6 - 2\, b) = 0[/tex].
[tex](7)\, (9 - a) + (-1)\, (3 - b) = 0[/tex].
[tex](-7)\, a + b = (-60)[/tex].
[tex]\left\lbrace \begin{aligned}& (-1)\, a + 3\, b = 20 \\ & (-7)\, a + b = (-60) \end{aligned} \right.[/tex].
[tex]a = 10[/tex].
[tex]b = 10[/tex].
Substitute the value of [tex]\text{$a$ and $b$}[/tex] back into the original equation (e.g., [tex](10 - a)^{2} + (20 - b)^{2} = r^{2}[/tex]) to find the value of [tex]r[/tex]:
[tex]r = 10[/tex].
In other words, the radius of the circle would be [tex]10[/tex], and the center should be at [tex](10,\, 10)[/tex].
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