Question

3 find the equation of the following: (1) a circle whose center is c (4, 2) that goes through p (5, 1). (2) a circle whose diameter is \\(\overline{ab}\\), a (-3, 1), and b (7, -9). 4 given that circle c : \\(x^2 + y^2 - 4x + 6y - 3 = 0\\), the center and the radius of the circle c are ______ , respectively. a. (-2, 3), 16 b. (2, -3), 4 c. (-2, 3), 4 d. (2, -3), 16

Explanation:

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Step1: Recall the circle equation formula

The standard equation of a circle with center \((h,k)\) and radius \(r\) is \((x - h)^2+(y - k)^2=r^2\). Here, the center \(C(4,2)\), so \(h = 4\), \(k = 2\).

Step2: Calculate the radius

The radius \(r\) is the distance between the center \(C(4,2)\) and the point \(P(5,1)\) on the circle. Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), we have \(r=\sqrt{(5 - 4)^2+(1 - 2)^2}=\sqrt{1 + 1}=\sqrt{2}\).

Step3: Write the circle equation

Substitute \(h = 4\), \(k = 2\), and \(r=\sqrt{2}\) into the standard equation: \((x - 4)^2+(y - 2)^2=(\sqrt{2})^2=2\). Expanding, we get \(x^{2}-8x + 16+y^{2}-4y + 4 = 2\), or \(x^{2}+y^{2}-8x-4y + 18 = 0\).

Step1: Find the center of the circle

The center of the circle is the midpoint of the diameter \(\overline{AB}\). The midpoint formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\). For \(A(-3,1)\) and \(B(7,-9)\), the center \((h,k)=(\frac{-3 + 7}{2},\frac{1+(-9)}{2})=(2,-4)\).

Step2: Calculate the radius

The radius \(r\) is half the length of \(\overline{AB}\). First, find the length of \(\overline{AB}\) using the distance formula: \(d=\sqrt{(7-(-3))^2+(-9 - 1)^2}=\sqrt{100 + 100}=\sqrt{200}=10\sqrt{2}\). So \(r=\frac{10\sqrt{2}}{2}=5\sqrt{2}\).

Step3: Write the circle equation

Substitute \(h = 2\), \(k=-4\), and \(r = 5\sqrt{2}\) into the standard equation: \((x - 2)^2+(y + 4)^2=(5\sqrt{2})^2 = 50\). Expanding, we get \(x^{2}-4x + 4+y^{2}+8y + 16 = 50\), or \(x^{2}+y^{2}-4x + 8y-30 = 0\).

Step1: Recall the general circle equation form

The general equation of a circle is \(x^{2}+y^{2}+Dx + Ey+F = 0\), and the center is \((-\frac{D}{2},-\frac{E}{2})\), radius \(r=\frac{1}{2}\sqrt{D^{2}+E^{2}-4F}\). For the circle \(x^{2}+y^{2}-4x + 6y-3 = 0\), we have \(D=-4\), \(E = 6\), \(F=-3\).

Step2: Find the center

The center \((h,k)=(-\frac{-4}{2},-\frac{6}{2})=(2,-3)\).

Step3: Find the radius

Calculate \(r=\frac{1}{2}\sqrt{(-4)^{2}+6^{2}-4\times(-3)}=\frac{1}{2}\sqrt{16 + 36 + 12}=\frac{1}{2}\sqrt{64}=\frac{1}{2}\times8 = 4\).

Answer:

\((x - 4)^2+(y - 2)^2 = 2\) (or expanded form \(x^{2}+y^{2}-8x-4y + 18 = 0\))

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