Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Answer:
[tex](x-5)^2+(y-10)^2=65[/tex]
Step-by-step explanation:
The general equation of a circle is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{General equation of a circle}}\\\\(x-h)^2+(y-k)^2=r^2\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\textsf{$(h, k)$ is the center.}\\\phantom{ww}\bullet\;\textsf{$r$ is the radius.}\end{array}}[/tex]
Given that the circle passes through the points (1, 3), (-2, 6), and (4, 2), we can substitute the coordinates of each point into the general equation to create a system of three equations:
[tex](1-h)^2+(3-k)^2=r^2\\\\(-2-h)^2+(6-k)^2=r^2\\\\(4-h)^2+(2-k)^2=r^2[/tex]
Set the first equation equal to the second equation:
[tex](1-h)^2+(3-k)^2=(-2-h)^2+(6-k)^2[/tex]
Solve for k:
[tex]1-2h+h^2+9-6k+k^2=4+4h+h^2+36-12k+k^2 \\\\ -2h+10-6k=4h+40-12k \\\\ 6k=6h+30\\\\k=h+5[/tex]
Set the second equation equal to the third equation:
[tex](-2-h)^2+(6-k)^2=(4-h)^2+(2-k)^2[/tex]
Solve for k:
[tex]4+4h+h^2+36-12k+k^2=16-8h+h^2+4-4k+k^2 \\\\4h+40-12k=20-8h-4k \\\\-8k=-12h-20\\\\8k=12h+20\\\\2k=3h+5\\\\k=\dfrac{3h+5}{2}[/tex]
Set the two equations for k equal to each other, and solve for h:
[tex]\dfrac{3h+5}{2}=h+5 \\\\3h+5=2h+10\\\\h = 5[/tex]
Substitute h = 5 into one of the equations for k, and solve for k:
[tex]k = h + 5 \\\\ k = 5 + 5 \\\\ k = 10[/tex]
Now, substitute h = 5 and k = 10 into one of the equations and solve for r²:
[tex](1-h)^2+(3-k)^2=r^2\\\\(1-5)^2+(3-10)^2=r^2 \\\\(-4)^2+(-7)^2=r^2\\\\16+49=r^2\\\\r^2=65[/tex]
Finally, substitute h = 5, k = 10 and r² = 65 into the general equation of a circle:
[tex](x-5)^2+(y-10)^2=65[/tex]
Therefore, the equation of the circle that passes through the points (1, 3), (-2, 6), and (4, 2) is:
[tex]\Large\boxed{\boxed{(x-5)^2+(y-10)^2=65}}[/tex]
This circle has a center located at point (5, 10) and a radius of [tex]\sqrt{65}[/tex].
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.