Question

what is the length of the radius of a circle with the general form equation $x^{2}+y^{2}-4x + 6y + 4 = 0$?
the equation of a circle is $(x + 3)^{2}+(y - 5)^{2}=16$. what is the distance from the center to the point $(1,5)$?

Explanation:

Step1: Find the radius of the first circle

The general form of a circle equation is \(x^{2}+y^{2}+Dx + Ey+F = 0\), and the radius \(r=\sqrt{\frac{D^{2}+E^{2}-4F}{4}}\).
For the equation \(x^{2}+y^{2}-4x + 6y+4 = 0\), where \(D=-4\), \(E = 6\), \(F = 4\).

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Step2: Find the distance for the second circle

The standard form of a circle equation is \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center.
For the equation \((x + 3)^{2}+(y - 5)^{2}=16\), the center is \((-3,5)\).
The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}\).
Here \((x_1,y_1)=(-3,5)\), \((x_2,y_2)=(1,5)\)

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Answer:

For the first question, the radius is \(3\). For the second question, the distance is \(4\).