
If the equation \[\begin{array}{*{20}{c}}
{p{x^2} + (2 - q)xy + 3{y^2} - 6qx + 30y + 6q}& = &0
\end{array}\] represents a circle, then the values of p and q are,
A) 2, 2
B) 3, 1
C) 3, 2
D) 3, 4
Answer
163.2k+ views
Hint: In this problem, we will compare the given equation with the general equation of the circle and we will our required answer
Complete Step by step solution:
We know the homogeneous equation that is written as,
\[\begin{array}{*{20}{c}}
{ \Rightarrow a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c}& = &0
\end{array}\]
Homogenous equations are the functions of the two variables. But if the coefficient of the \[xy\]is zero and the coefficient of \[{x^2}\] and \[{y^2}\] is the same, then the converted form of this equation is called the equation of the circle.
Now we know the general equation of the circle that is written as,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}}& = &{{r^2}}
\end{array}\]
And this equation may be evaluated as,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x^2} + {a^2} - 2ax + {y^2} + {b^2} - 2by}& = &0
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x^2} + {y^2} - 2ax - 2by + {a^2} + {b^2}}& = &0
\end{array}\] ………….. (A)
Now according to the question, we have given the equation that represents the circle. That is why we will compare that given equation with the equation (A).
Therefore, we will write
\[\begin{array}{*{20}{c}}
{ \Rightarrow p{x^2} + (2 - q)xy + 3{y^2} - 6qx + 30y + 6q}& = &0
\end{array}\]
We know that the coefficient of \[xy\]is zero for the circle. And the coefficients of \[{x^2}\]and \[{y^2}\]are the same for the circle. Therefore,
When we compare the above equation with equation (A). we will get,
\[\begin{array}{*{20}{c}}
{ \Rightarrow p}& = &3
\end{array}\]
And
\[\begin{array}{*{20}{c}}
{ \Rightarrow 2 - q}& = &0
\end{array}\] (because coefficient of \[xy\]is zero for the circle)
\[\begin{array}{*{20}{c}}
{ \Rightarrow q}& = &2
\end{array}\]
Now the values of the p and q are 1 and 2 respectively.
So the correct answer is option(C).
Note: The first point to keep in mind is that the coefficients of the \[xy\] are zero for the circle, and the coefficients of \[{x^2}\] and \[{y^2}\] are the same for the circle. The equation of the circle is formed only when all the conditions of the circle are satisfied in the homogeneous equation.
Complete Step by step solution:
We know the homogeneous equation that is written as,
\[\begin{array}{*{20}{c}}
{ \Rightarrow a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c}& = &0
\end{array}\]
Homogenous equations are the functions of the two variables. But if the coefficient of the \[xy\]is zero and the coefficient of \[{x^2}\] and \[{y^2}\] is the same, then the converted form of this equation is called the equation of the circle.
Now we know the general equation of the circle that is written as,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {{\left( {x - a} \right)}^2} + {{\left( {y - b} \right)}^2}}& = &{{r^2}}
\end{array}\]
And this equation may be evaluated as,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x^2} + {a^2} - 2ax + {y^2} + {b^2} - 2by}& = &0
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow {x^2} + {y^2} - 2ax - 2by + {a^2} + {b^2}}& = &0
\end{array}\] ………….. (A)
Now according to the question, we have given the equation that represents the circle. That is why we will compare that given equation with the equation (A).
Therefore, we will write
\[\begin{array}{*{20}{c}}
{ \Rightarrow p{x^2} + (2 - q)xy + 3{y^2} - 6qx + 30y + 6q}& = &0
\end{array}\]
We know that the coefficient of \[xy\]is zero for the circle. And the coefficients of \[{x^2}\]and \[{y^2}\]are the same for the circle. Therefore,
When we compare the above equation with equation (A). we will get,
\[\begin{array}{*{20}{c}}
{ \Rightarrow p}& = &3
\end{array}\]
And
\[\begin{array}{*{20}{c}}
{ \Rightarrow 2 - q}& = &0
\end{array}\] (because coefficient of \[xy\]is zero for the circle)
\[\begin{array}{*{20}{c}}
{ \Rightarrow q}& = &2
\end{array}\]
Now the values of the p and q are 1 and 2 respectively.
So the correct answer is option(C).
Note: The first point to keep in mind is that the coefficients of the \[xy\] are zero for the circle, and the coefficients of \[{x^2}\] and \[{y^2}\] are the same for the circle. The equation of the circle is formed only when all the conditions of the circle are satisfied in the homogeneous equation.
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