Answer
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Hint: The general form of the circle is given by the equation, \[{x^2} + {y^2} + 2gx + 2fy + c = 0\].
The radius of the circle is given by, \[{g^2} + {f^2} + c\]and the center of the circle is given by, \[( - g, - f)\].
Any point that passes through the circle will satisfy the equation of the circle, so we will substitute the given points and get two equations in terms of \[g\& f\]. Also, \[( - g, - f)\] is the center so it will satisfy the new equations so substituting these in that equation we will obtain the values of \[g\& f\].Then by substituting the values in the general equation we will get the required circle equation.
Complete step-by-step solution:
We know that the general form of the circle is given by, \[{x^2} + {y^2} + 2gx + 2fy + c = 0\], where \[( - g, - f)\]is the center of the circle.
It is given that this circle is passing through the points \[(1, - 2)\]and \[(4, - 3)\]. Thus, these points will satisfy the equation of the circle.
Let us substitute these points in the equation of the circle.
The equation of the circle at the point \[(1, - 2)\]is given by,
\[1 + 4 + 2g - 4f + c = 0\]
Simplifying this we get,
\[5 + 2g-4f + c = 0 …………….(1)\]
The equation of the circle at the point \[(4, - 3)\] is given by,
\[16 + 9 + 8g - 6f + c = 0\]
Simplifying the above equation, we get
\[25 + 8g - 6f + c = 0..........(2)\]
Solving \[(1)\] and \[(2)\],
Subtracting equation \[(1)\] from \[(2)\], we get
\[25 + 8g - 6f + c - [5 + 2g - 4f + c] =0\]
\[\Rightarrow 6g-2f+20 =0\]
\[ \Rightarrow 6g - 2f = - 20...........(3)\]
We know that, \[( - g, - f)\] is the center of the circle which will pass through the line \[3x + 4y = 7\].
Let us substitute the center in this line equation, we get
\[ - 3g - 4f = 7............(4)\]
Let us now solve \[(3)\] and \[(4)\] to find the values of \[g\& f\].
Multiply equation \[(4)\] by 2 we get, \[ - 6g - 8f = 14\].
Now adding equation \[(3)\] to the above equation we get,
\[ - 10f = - 6\]
\[ \Rightarrow f = \dfrac{3}{5}\]
Substituting the value of \[f\] in equation \[(4)\] we get,
\[
- 3g - 4(\dfrac{3}{5}) = 7 \\
\Rightarrow - 3g - \dfrac{{12}}{5} = 7 \\
\Rightarrow - 3g = 7 + \dfrac{{12}}{5} \\
\Rightarrow - 3g = \dfrac{{47}}{5} \\
\Rightarrow g = \dfrac{{ - 47}}{{15}} \]
Thus, we got the values of \[g = \dfrac{{ - 47}}{{15}}\& f = \dfrac{3}{5}\]. Let us substitute these values in the equation \[(1)\] we get,
\[
5 + 2\left( {\dfrac{{ - 47}}{{15}}} \right) - 4\left( {\dfrac{3}{5}} \right) + c = 0 \\
\Rightarrow 5 - \dfrac{{94}}{{15}} - \dfrac{{12}}{5} + c = 0 \\
\Rightarrow c = \dfrac{{55}}{15} \]
Now let us substitute the values of \[f,g\& c\]in the circle equation we get,
\[ {x^2} + {y^2} - \dfrac{{94}}{{15}}x + \dfrac{{18}}{{15}}y + \dfrac{{55}}{{15}} = 0 \\
\Rightarrow 15{x^2} + 15{y^2} - 94x + 18y + 55 = 0 \]
Thus, this is the required circle equation.
Note: In this problem it is given that the circle is passing through two points and the center of the circle is passing through the line, this helped us to find the equation of the circle coordinates. After finding the values of circle coordinates we will substitute it in the circle equation to get the required circle.
The radius of the circle is given by, \[{g^2} + {f^2} + c\]and the center of the circle is given by, \[( - g, - f)\].
Any point that passes through the circle will satisfy the equation of the circle, so we will substitute the given points and get two equations in terms of \[g\& f\]. Also, \[( - g, - f)\] is the center so it will satisfy the new equations so substituting these in that equation we will obtain the values of \[g\& f\].Then by substituting the values in the general equation we will get the required circle equation.
Complete step-by-step solution:
We know that the general form of the circle is given by, \[{x^2} + {y^2} + 2gx + 2fy + c = 0\], where \[( - g, - f)\]is the center of the circle.
It is given that this circle is passing through the points \[(1, - 2)\]and \[(4, - 3)\]. Thus, these points will satisfy the equation of the circle.
Let us substitute these points in the equation of the circle.
The equation of the circle at the point \[(1, - 2)\]is given by,
\[1 + 4 + 2g - 4f + c = 0\]
Simplifying this we get,
\[5 + 2g-4f + c = 0 …………….(1)\]
The equation of the circle at the point \[(4, - 3)\] is given by,
\[16 + 9 + 8g - 6f + c = 0\]
Simplifying the above equation, we get
\[25 + 8g - 6f + c = 0..........(2)\]
Solving \[(1)\] and \[(2)\],
Subtracting equation \[(1)\] from \[(2)\], we get
\[25 + 8g - 6f + c - [5 + 2g - 4f + c] =0\]
\[\Rightarrow 6g-2f+20 =0\]
\[ \Rightarrow 6g - 2f = - 20...........(3)\]
We know that, \[( - g, - f)\] is the center of the circle which will pass through the line \[3x + 4y = 7\].
Let us substitute the center in this line equation, we get
\[ - 3g - 4f = 7............(4)\]
Let us now solve \[(3)\] and \[(4)\] to find the values of \[g\& f\].
Multiply equation \[(4)\] by 2 we get, \[ - 6g - 8f = 14\].
Now adding equation \[(3)\] to the above equation we get,
\[ - 10f = - 6\]
\[ \Rightarrow f = \dfrac{3}{5}\]
Substituting the value of \[f\] in equation \[(4)\] we get,
\[
- 3g - 4(\dfrac{3}{5}) = 7 \\
\Rightarrow - 3g - \dfrac{{12}}{5} = 7 \\
\Rightarrow - 3g = 7 + \dfrac{{12}}{5} \\
\Rightarrow - 3g = \dfrac{{47}}{5} \\
\Rightarrow g = \dfrac{{ - 47}}{{15}} \]
Thus, we got the values of \[g = \dfrac{{ - 47}}{{15}}\& f = \dfrac{3}{5}\]. Let us substitute these values in the equation \[(1)\] we get,
\[
5 + 2\left( {\dfrac{{ - 47}}{{15}}} \right) - 4\left( {\dfrac{3}{5}} \right) + c = 0 \\
\Rightarrow 5 - \dfrac{{94}}{{15}} - \dfrac{{12}}{5} + c = 0 \\
\Rightarrow c = \dfrac{{55}}{15} \]
Now let us substitute the values of \[f,g\& c\]in the circle equation we get,
\[ {x^2} + {y^2} - \dfrac{{94}}{{15}}x + \dfrac{{18}}{{15}}y + \dfrac{{55}}{{15}} = 0 \\
\Rightarrow 15{x^2} + 15{y^2} - 94x + 18y + 55 = 0 \]
Thus, this is the required circle equation.
Note: In this problem it is given that the circle is passing through two points and the center of the circle is passing through the line, this helped us to find the equation of the circle coordinates. After finding the values of circle coordinates we will substitute it in the circle equation to get the required circle.
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