-Welp-
  • -Welp-
"What is the equation of the circle with center (3,5) that passes through the point (-4,10)?" ^How do I solve this?
Geometry
katieb
  • katieb
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anonymous
  • anonymous
Use the distance formula first to find the radius.\[\huge d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\] After you found the radius, use the center and the radius to form the standard equation of a circle.
Nnesha
  • Nnesha
or you can use equation of the circle replace (h,k) for center points and x , y by the point they gave you solve for r
jdoe0001
  • jdoe0001
|dw:1438730789405:dw|

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jdoe0001
  • jdoe0001
\(\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({\color{red}{ 3}}\quad ,&{\color{blue}{ 5}})\quad % (c,d) &({\color{red}{ -4}}\quad ,&{\color{blue}{ 10}}) \end{array}\qquad % distance value d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)
jdoe0001
  • jdoe0001
find the radius use the center (h,k) as Nnesha indicated on the previous posting and plug them in \(\bf (x-{\color{brown}{ h}})^2+(y-{\color{blue}{ k}})^2={\color{purple}{ r}}^2 \qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})\qquad radius={\color{purple}{ r}}\)
zzr0ck3r
  • zzr0ck3r
@Nnesha solving for r in the standard equation gives the distance formula :)
Nnesha
  • Nnesha
\[\huge\rm (\color{blue}{x}-\color{reD}{h})^2+(\color{blue}{y}-\color{reD}{k})^2=r^2\] \[\color{blue}{(-4,10)}\] \[\color{reD}{(3,5)}\] plugin values solve for r
Nnesha
  • Nnesha
yeah same thing
Nnesha
  • Nnesha
Opps nvm
-Welp-
  • -Welp-
"plugin values solve for r " I tried it and got 24. That doesn't seem right.
Nnesha
  • Nnesha
wrong.
-Welp-
  • -Welp-
=(x-3)^2 + (y-5)^2=226?
-Welp-
  • -Welp-
halp
anonymous
  • anonymous
The left hand side of your equation of the circle is correct. However, the right hand side (226) is incorrect. Did you use the distance formula above to find the radius?

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