Equation of a Circle Given Two Points and Tangent Line

Consider the general equation circle is given by

{x^2}  + {y^2} + 2gx + 2fy + c = 0\,\,\,{\text{ -   -  - }}\left( {\text{i}} \right)


two-points-tangent-line-circle

If the given circle is passing through two points, say, A\left(  {{x_1},{y_1}} \right)and B\left(  {{x_2},{y_2}} \right), then these points must satisfy the general equation of circle. Now put these two points in the given equation of circle, i.e.

\begin{gathered} {x_1}^2 + {y_1}^2 + 2g{x_1} + 2f{y_1} + c =  0\,\,\,{\text{ -  -  - }}\,\left( {{\text{ii}}} \right) \\ {x_2}^2 + {y_2}^2 + 2g{x_2} + 2f{y_2} + c =  0\,\,\,{\text{ -  -  - }}\,\left( {{\text{iii}}} \right) \\ \end{gathered}


Also the given straight line ax + by + d = 0\,\,\,{\text{ -  -  -  }}\left( {{\text{iv}}} \right) touches the circle at one point as shown in the given diagram and it is clear from the diagram that the distance of a given form the center \left( { - g, - f}  \right) must be equal to the radius of the circle. By using the formula for distance between point and the line i.e.

r =  \frac{{\left| { - ag - bf + d} \right|}}{{\sqrt {{a^2} + {b^2}} }}


But the radius of the circle in general form is r = \sqrt {{g^2} + {f^2} - c} , using this values we get

  \Rightarrow \sqrt {{g^2} + {f^2} - c}  =  \frac{{\left| { - ag - bf + d} \right|}}{{\sqrt {{a^2} + {b^2}} }}


Squaring both sides of the above equation we have

 \Rightarrow  {g^2} + {f^2} - c = \frac{{{{\left( { - ag - bf + d} \right)}^2}}}{{{a^2} +  {b^2}}}\,\,\,{\text{ -  -  - }}\left( {\text{v}} \right)


To evaluate the equation of required circle, we must the find values of g,f,c from the above equations (ii), (iii) and (v). We can solve these three using the method of simultaneous equations, then put all these in the equation (i) to get the required circle.
An alternate way is that we assume that the required equation of circle is

{\left(  {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\,\,\,{\text{ -  -  -  }}\left( {\text{i}} \right)


Since the circle passes through the points A\left( {{x_1},{y_1}} \right)and B\left( {{x_2},{y_2}} \right), so these points must satisfy the above equation of circle, we have

\begin{gathered} {\left( {{x_1} - h} \right)^2} + {\left(  {{y_1} - k} \right)^2} = {r^2}\,\,\,{\text{ -   -  - }}\left( {{\text{ii}}}  \right) \\ {\left( {{x_2} - h} \right)^2} + {\left(  {{y_2} - k} \right)^2} = {r^2}\,\,\,{\text{ -   -  - }}\left( {{\text{iii}}}  \right) \\ \end{gathered}


Since the given line ax  + by + d = 0\,\,\,{\text{ -  -  - }}\left( {{\text{iv}}} \right) touches the circle.
So the distance of the center of the circle from the given line ax + by + d = 0 must be equal to its radius, i.e.

r =  \frac{{\left| {a{x_1} + b{y_1} + d} \right|}}{{\sqrt {{a^2} + {b^2}}  }}\,\,\,{\text{ -  -  - }}\left( {\text{v}} \right)


Now using the above equation (ii), (iii) and (v), we can find the values of h,k and r in the equation (i), we get the required circle.

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