Question Statement

Find the equation of the circle passing through the point $(2,1)$ and touching the line $x+2y-1=0$ at the point $(3,-1)$.

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Background and Explanation

To solve this problem, we use the property that the distance from the center of a circle to any point on its circumference is equal (the radius). Additionally, the radius of a circle is always perpendicular to the tangent line at the point of contact. By setting up equations based on these geometric properties, we can find the coordinates of the center $(h,k)$ and the radius $r$.

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Solution

Let $C(h,k)$ be the center of the circle. Let $A(2,1)$ be the point the circle passes through, and $B(3,-1)$ be the point where the circle touches the given line.

1. Equating the Radii

Since both $A$ and $B$ lie on the circle, the distance from the center $C$ to $A$ must equal the distance from $C$ to $B$ ($|AC|=|BC|$):

$\sqrt{(h-2{)}^2+(k-1{)}^2}=\sqrt{(h-3{)}^2+(k+1{)}^2}$

Squaring both sides to remove the radicals:

$(h-2{)}^2+(k-1{)}^2=(h-3{)}^2+(k+1{)}^2$

Expanding the squares:

$h^2-4h+4+k^2-2k+1=h^2-6h+9+k^2+2k+1$

Subtracting $h^2$ and $k^2$ from both sides and simplifying:

$-4h-2k+5=-6h+2k+10$ $2h-4k=5$

2. Using the Tangent Property

The slope of the given tangent line $x+2y-1=0$ is $m_1=-\frac{1}{2}$. Since the radius $\overline{BC}$ is perpendicular to the tangent line at point $B$, the product of their slopes must be $-1$.

The slope of $\overline{BC}$ is $m_2=\frac{k-(-1)}{h-3}=\frac{k+1}{h-3}$.

$(m_2)(m_1)=-1\Rightarrow \left(\frac{k+1}{h-3}\right)\left(\frac{-1}{2}\right)=-1$

Multiplying both sides by $-2$:

$\frac{k+1}{h-3}=2$ $k+1=2(h-3)$ $k+1=2h-6$ $2h-k=7$

3. Solving for the Center $(h,k)$

We now have a system of two linear equations:

1. $2h-4k=5$
2. $2h-k=7$

Subtracting equation (ii) from equation (i):

$\begin{array}{rl}(2h-4k)-(2h-k)& =5-7\\ -3k& =-2\\ k& =\frac{2}{3}\end{array}$

Substitute $k=\frac{2}{3}$ into equation (ii):

$\begin{array}{rl}2h-\frac{2}{3}& =7\\ 2h& =7+\frac{2}{3}\\ 2h& =\frac{23}{3}\\ h& =\frac{23}{6}\end{array}$

Thus, the center of the circle is $C\left(\frac{23}{6},\frac{2}{3}\right)$.

4. Finding the Radius

The radius squared ($r^2$) is the distance $|AC{|}^2$:

$\begin{array}{rl}{r}^2& ={\left(\frac{23}{6}-2\right)}^2+{\left(\frac{2}{3}-1\right)}^2\\ & ={\left(\frac{11}{6}\right)}^2+{\left(\frac{-1}{3}\right)}^2\\ & =\frac{121}{36}+\frac{1}{9}=\frac{121+4}{36}=\frac{125}{36}\end{array}$

5. Equation of the Circle

Using the standard form $(x-h{)}^2+(y-k{)}^2=r^2$:

${\left(x-\frac{23}{6}\right)}^2+{\left(y-\frac{2}{3}\right)}^2=\frac{125}{36}$

Expanding the terms:

$x^2-\frac{23}{3}x+\frac{529}{36}+y^2-\frac{4}{3}y+\frac{4}{9}=\frac{125}{36}$

Rearranging and combining constants:

$x^2+y^2-\frac{23}{3}x-\frac{4}{3}y+\left(\frac{529}{36}+\frac{16}{36}-\frac{125}{36}\right)=0$ $x^2+y^2-\frac{23}{3}x-\frac{4}{3}y+\frac{420}{36}=0$ $x^2+y^2-\frac{23}{3}x-\frac{4}{3}y+\frac{35}{3}=0$

Multiply the entire equation by 3 to clear the denominators:

$3x^2+3y^2-23x-4y+35=0$

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Key Formulas or Methods Used

• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$
• Perpendicular Slopes: $m_1\cdot m_2=-1$
• Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

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Summary of Steps

1. Define the center as $(h,k)$ and set the distance from the center to $(2,1)$ equal to the distance to $(3,-1)$.
2. Determine the slope of the tangent line and use the perpendicularity condition to find the slope of the radius.
3. Solve the resulting system of two linear equations to find the center coordinates $(h,k)$.
4. Calculate the radius squared using the distance formula between the center and one of the points.
5. Substitute the center and radius into the circle equation and simplify to the general form.