Question

question 14
2.86 pts
find an equation of a circle that satisfies the following condition. write your answer in standard form.
center: (7, 1); passing through (-5, 3)
$(x - 7)^2+(y + 1)^2=(\sqrt{37})^2$
$(x - 7)^2+(y - 1)^2=(2\sqrt{37})^2$
$(x + 5)^2+(y - 3)^2=(\sqrt{34})^2$
$(x - 7)^2+(y - 1)^2=(\sqrt{34})^2$
$(x + 5)^2+(y - 3)^2=(\sqrt{37})^2$

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius. Given the center \((7,1)\), the equation becomes \((x - 7)^2+(y - 1)^2=r^2\).

Step2: Calculate the radius

Use the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) to find the radius. Here, \((x_1,y_1)=(7,1)\) and \((x_2,y_2)=(-5,3)\).

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Step3: Substitute the radius into the equation

Substitute \(r = 2\sqrt{37}\) into \((x - 7)^2+(y - 1)^2=r^2\), we get \((x - 7)^2+(y - 1)^2=(2\sqrt{37})^2\).

Answer:

\(\boldsymbol{(x - 7)^2+(y - 1)^2=(2\sqrt{37})^2}\) (the second option)