Question

5 find the equation of the circle that passes through a(1, -1), b(1, 4), and c(4, -2), and find the radius and the center of the circle.

6 the center of a circle is (3, 7) and the circle passes through (2, 5). the equation of this circle is ______ .

Explanation:

Problem 5 (Finding circle equation through three points)

Step1: Recall circle equation

The general equation of a circle is \((x - h)^2+(y - k)^2 = r^2\), where \((h,k)\) is the center and \(r\) is the radius. Since the circle passes through \(A(1,-1)\), \(B(1,4)\), and \(C(4,-2)\), we substitute these points into the equation.

For point \(A(1,-1)\): \((1 - h)^2+(-1 - k)^2=r^2\) --- (1)

For point \(B(1,4)\): \((1 - h)^2+(4 - k)^2=r^2\) --- (2)

For point \(C(4,-2)\): \((4 - h)^2+(-2 - k)^2=r^2\) --- (3)

Step2: Subtract (1) from (2)

\((1 - h)^2+(4 - k)^2-( (1 - h)^2+(-1 - k)^2 )=r^2 - r^2\)

Simplify: \((4 - k)^2-(-1 - k)^2 = 0\)

Expand: \((16 - 8k + k^2)-(1 + 2k + k^2)=0\)

Simplify: \(16 - 8k + k^2 - 1 - 2k - k^2 = 0\)

\(15 - 10k = 0\) → \(10k = 15\) → \(k=\frac{3}{2}\)

Step3: Subtract (1) from (3)

\((4 - h)^2+(-2 - k)^2-( (1 - h)^2+(-1 - k)^2 )=r^2 - r^2\)

Expand \((4 - h)^2=16 - 8h + h^2\), \((1 - h)^2=1 - 2h + h^2\), \((-2 - k)^2=4 + 4k + k^2\), \((-1 - k)^2=1 + 2k + k^2\)

So: \((16 - 8h + h^2)+(4 + 4k + k^2)-(1 - 2h + h^2)-(1 + 2k + k^2)=0\)

Simplify: \(16 - 8h + h^2 + 4 + 4k + k^2 - 1 + 2h - h^2 - 1 - 2k - k^2 = 0\)

\(18 - 6h + 2k = 0\)

We know \(k=\frac{3}{2}\), substitute: \(18 - 6h + 2\times\frac{3}{2}=0\) → \(18 - 6h + 3 = 0\) → \(21 - 6h = 0\) → \(6h = 21\) → \(h=\frac{7}{2}\)

Step4: Find radius \(r\)

Substitute \(h=\frac{7}{2}\), \(k=\frac{3}{2}\) into equation (1):

\((1-\frac{7}{2})^2+(-1-\frac{3}{2})^2=r^2\)

\((-\frac{5}{2})^2+(-\frac{5}{2})^2=r^2\) → \(\frac{25}{4}+\frac{25}{4}=r^2\) → \(\frac{50}{4}=\frac{25}{2}=r^2\) → \(r=\frac{5\sqrt{2}}{2}\) (radius is positive)

Step5: Write circle equation

The center is \((\frac{7}{2},\frac{3}{2})\) and \(r^2=\frac{25}{2}\), so the equation is \((x - \frac{7}{2})^2+(y - \frac{3}{2})^2=\frac{25}{2}\)

We can also expand it: \(x^2 - 7x + \frac{49}{4}+y^2 - 3y + \frac{9}{4}=\frac{25}{2}\)

Multiply through by 4: \(4x^2 - 28x + 49 + 4y^2 - 12y + 9 = 50\)

\(4x^2 + 4y^2 - 28x - 12y - 2 = 0\) (or divide by 2: \(2x^2 + 2y^2 - 14x - 6y - 1 = 0\))

Problem 6 (Finding circle equation with center and a point)

Step1: Recall circle equation

The standard equation of a circle is \((x - h)^2+(y - k)^2 = r^2\), where \((h,k)=(3,7)\) is the center.

Step2: Find radius \(r\)

The circle passes through \((2,5)\), so the distance between \((3,7)\) and \((2,5)\) is the radius.

Using distance formula \(r=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)

\(r=\sqrt{(2 - 3)^2+(5 - 7)^2}=\sqrt{(-1)^2+(-2)^2}=\sqrt{1 + 4}=\sqrt{5}\)

So \(r^2 = 5\)

Step3: Write circle equation

Substitute \(h = 3\), \(k = 7\), \(r^2 = 5\) into the equation: \((x - 3)^2+(y - 7)^2 = 5\)

We can expand it: \(x^2 - 6x + 9 + y^2 - 14y + 49 = 5\) → \(x^2 + y^2 - 6x - 14y + 53 = 0\)

Answer:

s:

Problem 5:

• Equation of the circle: \(\boldsymbol{(x - \frac{7}{2})^2+(y - \frac{3}{2})^2=\frac{25}{2}}\) (or expanded form)
• Center: \(\boldsymbol{(\frac{7}{2},\frac{3}{2})}\)
• Radius: \(\boldsymbol{\frac{5\sqrt{2}}{2}}\)

Problem 6:

Equation of the circle: \(\boldsymbol{(x - 3)^2+(y - 7)^2 = 5}\) (or expanded form \(x^2 + y^2 - 6x - 14y + 53 = 0\))