How to find the equation of a circle if the givens are the:
Case 1: Tangent to $2x + 3y + 13 = 0$ and $2x - 3y - 1 = 0$; contains $(0,4)$
Case 2: Tangent to $x - 3y - 7 = 0$ and $3x + y - 21 = 0$; center on $x - 3y + 3 = 0$
Case 3: Tangent to $x - y = 0$ at $(2,2)$; center on $2x + 3y - 7 = 0$
Case 4: Tangent to $x^2 + y^2 - 22x + 20y + 77 = 0$ at $(91/17,10/17)$; containing $(0,1)$
Please help, I have no idea how to solve this problems. I've been stuck on these like forever.
Thank You
On
For case 1. Warning: this answer was given to the revision 1 of the question. In the present revision 2 the circle contains the point $(0,4)$ instead of the point $(-2,-1)$.
I detail the computation as follows. Let the equation of the circle be \begin{equation} \left( x-h\right) ^{2}+\left( y-k\right) ^{2}=r^{2}, \tag{1} \end{equation} where $(h,k)$ is the center and $r$ the radius. The slopes of the tangent lines
\begin{eqnarray} 2x+3y+13 &=&0 \tag{2} \\ && \\ 2x-3y-1 &=&0 \tag{3} \end{eqnarray}
are respectively $m_{1}=-2/3$ and $m_{2}=2/3$. Solving this system of equations $(2)$-$(3)$ we get $(x,y)=(-3,-7/3)$. Since the circle contains $P(-2,-1)$ its center $C(h,k)$ should lie on the (symmetric) vertical line $x=-3$, which means that $h=-3$ and $r=\overline{CP}$:
\begin{equation} r^{2}=(\overline{CP})^{2}=1+\left( -1-k\right) ^{2}=1+\left( 1+k\right) ^{2}. \tag{4} \end{equation}
$$\text{Circle (red)} \left( x+3\right) ^{2}+( y-2-\frac{1}{2}\sqrt{39}) ^{2}=\frac{79}{4}+3\sqrt{39}$$
$$ \text{tangent to lines (2),(3) and containing point} (-2,-1)$$
Let $Q(s,t)$ be the point where the circle touches the line $(3)$. Then
\begin{equation*} 2s-3t-1=0\Leftrightarrow t=\frac{1}{3}\left( 2s-1\right) . \end{equation*}
The slope $m$ of a line orthogonal to this tangent line is $m=-1/m_{2}=-3/2$; from this family of lines the equation of the one containing $Q$ is
\begin{equation*} y-t=-\frac{3}{2}(x-s). \end{equation*}
This line should intercept the vertical line $x=h=-3$ at $C$. Hence
\begin{equation*} k=-\frac{3}{2}(-3-s)+t=\frac{3}{2}(3+s)+\frac{1}{3}\left( 2s-1\right) =\frac{1}{6}(25+13s). \end{equation*}
To find $r$ we are going to equate $(\overline{CP})^{2}=(\overline{CQ})^{2}$. On the one hand, we have that \begin{eqnarray*} r^{2} &=&(\overline{CQ})^{2}=\left( s-h\right) ^{2}+\left( t-k\right) ^{2} \\ &=&(s+3)^{2}+(\frac{1}{3}\left( 2s-1\right) -\frac{1}{6}(25+13s))^{2} \\ &=&\frac{1}{4}(13s^{2}+78s+117).\tag{5} \end{eqnarray*} On the other hand \begin{eqnarray*} r^{2} &=&(\overline{CP})^{2}=1+(1+\frac{1}{6}(25+13s))^{2} \\ &=&\frac{1}{36}\left( 169s^{2}+806s+997\right)\tag{6} . \end{eqnarray*} So, equating $(5)$ and $(6)$ and simplifying yields \begin{equation*} 13s^{2}+26s-14=0\Leftrightarrow s_{1,2}=-1\pm \frac{3}{13}\sqrt{39}. \end{equation*}
For the positive solution $s_{1}$, we find that \begin{equation*} s_{1}=-1+\frac{3}{13}\sqrt{39}.\tag{7} \end{equation*}
Then
\begin{eqnarray} k &=&\frac{1}{6}(25+13s)=2+\frac{1}{2}\sqrt{39} \tag{8} \\ && \\ r^{2} &=&1+\left( 1+k\right) ^{2}=\frac{79}{4}+3\sqrt{39}. \tag{9} \end{eqnarray}
The equation of the circle is thus
\begin{equation} \left( x+3\right) ^{2}+\left( y-2-\frac{1}{2}\sqrt{39}\right) ^{2}=\frac{79}{4}+3\sqrt{39}. \tag{10} \end{equation}
ADDED. For the negative solution $s_2$, we have that
\begin{eqnarray*} s_{2} &=&-1-\frac{3}{13}\sqrt{39}\tag{11} \\ && \\ k &=&\frac{1}{6}(25+13s)=2-\frac{1}{2}\sqrt{39}\tag{12} \\ && \\ r^{2} &=&1+\left( 1+k\right) ^{2}=\frac{79}{4}-3\sqrt{39}.\tag{13} \end{eqnarray*}
The equation of this second circle (the green one in the figure below) is thus
\begin{equation*} \left( x+3\right) ^{2}+\left( y-2+\frac{1}{2}\sqrt{39}\right) ^{2}=\frac{79}{ 4}-3\sqrt{39}.\tag{14} \end{equation*}
$$ \text{The two circles: circle green } \left( x+3\right) ^{2}+\left( y-2+\frac{1}{2}\sqrt{39}\right) ^{2}=\frac{79}{ 4}-3\sqrt{39}$$
$$ \text{tangent to lines (2),(3) and containing point} (-2,-1)$$
For case 3. Let the equation of the circle be as above
\begin{equation} \left( x-h\right) ^{2}+\left( y-k\right) ^{2}=r^{2}. \tag{15} \end{equation}
The point $P(2,2)$ defines the condition
\begin{equation*} \left( 2-h\right) ^{2}+\left( 2-k\right) ^{2}=r^{2},\tag{16} \end{equation*} Since the line orthogonal to the tangent line $x-y=0$ containing $P(2,2)$ is given by the equation
\begin{equation*} y-2=-(x-2),\tag{17} \end{equation*}
the intersection of this line with the center line
\begin{equation*} 2x+3y-7=0\tag{18} \end{equation*}
is the center $C(h,k)=C(5,-1)$. So
\begin{equation*} r^{2}=\left( 2-h\right) ^{2}+\left( 2-k\right) ^{2}=\left( 2-5\right) ^{2}+\left( 2+1\right) ^{2}=18\tag{19}. \end{equation*}
The equation $(15)$ of the circle is thus
\begin{equation*} \left( x-5\right) ^{2}+\left( y+1\right) ^{2}=18.\tag{20} \end{equation*}
$$\text{Circle } \left( x-5\right) ^{2}+\left( y+1\right) ^{2}=18$$
For case 1 and case 2, the center lies on the bisector of the angle of the given lines
For case 3, can you write the equation for the line perpendicular to $x-y=0$ and going through $(2,2)$? Does that sound like the point-slope form? Now you have two simultaneous equations with $2x+3y-7=0$
For case 4, you need to take the derivative to find the slope of the tangent. Again you have a point-slope form.