Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Write the standard form of the equation of the circle with the given characteristics. Center: (3, 9); Solution point: (−2, 21)

Sagot :

Given:

Center of a circle is (3,9).

Solution point is (-2,21).

To find:

The standard form of the circle.

Solution:

Radius is the distance between center (3,9) and the solution point (-2,21).

[tex]r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]r=\sqrt{(-2-3)^2+(21-9)^2}[/tex]

[tex]r=\sqrt{(-5)^2+(12)^2}[/tex]

[tex]r=\sqrt{25+144}[/tex]

On further simplification, we get

[tex]r=\sqrt{169}[/tex]

[tex]r=13[/tex]

The radius of the circle is 13 units.

The standard form of a circle is

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Where, (h,k) is the center and r is the radius.

Putting h=3, k=9 and r=13, we get

[tex](x-3)^2+(y-9)^2=(13)^2[/tex]

[tex](x-3)^2+(y-9)^2=169[/tex]

Therefore, the standard form of the circle is [tex](x-3)^2+(y-9)^2=169[/tex].