Question

Find the equation of a circle having (1, −2) as its center and passing through the intersection of the lines 3x + y = 14 and 2x + 5y = 18.

Answer 1

Hint: First of all, determine the coordinates of the intersection point of the line 3x + y = 14 and 2x + 5y = 18. As we have given that the circle passes through the intersection point of the lines, the coordinates of that intersection point satisfy the equation of the circle. So, we will determine the radius of the circle. Hence, we will get our desired answer.

Complete Step by step solution:
According to the question, we have given that the center of the circle is (1, -2).
Suppose that R is the radius of the circle. So, we know that the general equation of the circle that is,
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}}& = &{{R^2}}
\end{array}\]
We have given the coordinates of the circle (1, -2). Where,
a = 1 and b = -2
Put these values in the above equation. We will get
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\left( {x - 1} \right)}^2} + {{\left( {y + 2} \right)}^2}}& = &{{R^2}}
\end{array}\]……………….. (A).
Now we will determine the coordinates of the intersection point of the lines. Therefore, we have given
3x + y = 14 -----(1)
And
2x + 5y = 18 -----(2)
Form the equation (1) and (2).
We will get the value of x and y. Therefore,
\[\begin{array}{*{20}{c}}
  { \Rightarrow x}& = &4
\end{array}\] and \[\begin{array}{*{20}{c}}
  y& = &2
\end{array}\]
Now the coordinates of the intersection point of the lines are (4, 2). Hence the circle passes through these points. So, these coordinates will satisfy the equation of the circle. Form the equation (A).
 \[\begin{array}{*{20}{c}}
  { \Rightarrow {{\left( {4 - 1} \right)}^2} + {{\left( {2 + 2} \right)}^2}}& = &{{R^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow 9 + 16}& = &{{R^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow R}& = &5
\end{array}\]
Now again form the equation (A).
\[\begin{array}{*{20}{c}}
  { \Rightarrow {{\left( {x - 1} \right)}^2} + {{\left( {y + 2} \right)}^2}}& = &{{{\left( 5 \right)}^2}}
\end{array}\]
\[\begin{array}{*{20}{c}}
  { \Rightarrow {x^2} + {y^2} - 2x + 4y - 20}& = &0
\end{array}\]

Note: It is important to note that If any circle passes through the point, then the coordinates of that point will satisfy the equation of the circle.