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Question: Find the equation of the circle which passes through the points \[(1, - 2)\] and \[(4, - 3)\], whose...

Find the equation of the circle which passes through the points (1,2)(1, - 2) and (4,3)(4, - 3), whose center lies on the line 3x+4y=73x + 4y = 7.

Explanation

Solution

The general form of the circle is given by the equation, x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0.
The radius of the circle is given by, g2+f2+c{g^2} + {f^2} + cand the center of the circle is given by, (g,f)( - 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&fg\& f. Also, (g,f)( - 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&fg\& 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, x2+y2+2gx+2fy+c=0{x^2} + {y^2} + 2gx + 2fy + c = 0, where (g,f)( - g, - f)is the center of the circle.
It is given that this circle is passing through the points (1,2)(1, - 2)and (4,3)(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)(1, - 2)is given by,
1+4+2g4f+c=01 + 4 + 2g - 4f + c = 0
Simplifying this we get,
5+2g4f+c=0.(1)5 + 2g-4f + c = 0 …………….(1)
The equation of the circle at the point (4,3)(4, - 3) is given by,
16+9+8g6f+c=016 + 9 + 8g - 6f + c = 0
Simplifying the above equation, we get
25+8g6f+c=0..........(2)25 + 8g - 6f + c = 0..........(2)
Solving (1)(1) and (2)(2),
Subtracting equation (1)(1) from (2)(2), we get
25+8g6f+c[5+2g4f+c]=025 + 8g - 6f + c - [5 + 2g - 4f + c] =0
6g2f+20=0\Rightarrow 6g-2f+20 =0
6g2f=20...........(3)\Rightarrow 6g - 2f = - 20...........(3)
We know that, (g,f)( - g, - f) is the center of the circle which will pass through the line 3x+4y=73x + 4y = 7.
Let us substitute the center in this line equation, we get
3g4f=7............(4)- 3g - 4f = 7............(4)
Let us now solve (3)(3) and (4)(4) to find the values of g&fg\& f.
Multiply equation (4)(4) by 2 we get, 6g8f=14 - 6g - 8f = 14.
Now adding equation (3)(3) to the above equation we get,
10f=6- 10f = - 6
f=35\Rightarrow f = \dfrac{3}{5}
Substituting the value of ff in equation (4)(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} Nowletussubstitutethevaluesof Now let us substitute the values off,g& c$$in the circle equation we get,

\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.