Question

what is the radius of a circle with the equation $(x - 5)^2+(y + 4)^2=25$?
a. 25
b. 5
c. 4
d. 6
which equation represents a circle passing through the point $(4,3)$ with a center at $(2,1)$?
a. $(x - 2)^2+(y - 1)^2=8$
b. $(x - 2)^2+(y - 1)^2=13$
c. $(x - 2)^2+(y - 1)^2=4$
d. $(x - 2)^2+(y - 1)^2=25$

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: For the first question, find the radius

Given the equation \((x - 5)^2+(y + 4)^2 = 25\). Comparing with \((x - h)^2+(y - k)^2=r^2\), we have \(r^2=25\), so \(r = 5\) (since \(r>0\)).

Step3: For the second question, find the radius of the circle passing through \((4,3)\) with center \((2,1)\)

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here \(x_1 = 2,y_1 = 1,x_2=4,y_2 = 3\). Then \(r=\sqrt{(4 - 2)^2+(3 - 1)^2}=\sqrt{4 + 4}=\sqrt{8}\). The equation of the circle is \((x - 2)^2+(y - 1)^2=8\) (using the standard form \((x - h)^2+(y - k)^2=r^2\) with \(h = 2,k = 1,r^2=8\)).

Answer:

First question: b. 5
Second question: a. \((x - 2)^2+(y - 1)^2=8\)