
A circle touches the x-axis and cuts off a chord of length 2l from the y-axis. The locus of the center of the circle is
A) A straight line
B) A circle
C) An ellipse
D) A hyperbola
Answer
164.7k+ views
Hint: In this question, we have given that the circle touches the x axis. Therefore, apply the formula \[\begin{array}{*{20}{c}}
d& = &{2\sqrt {{g^2} - c} }
\end{array}\]where d will be zero because the circle touches the x axis. And then apply the formula \[\begin{array}{*{20}{c}}
d& = &{2\sqrt {{f^2} - c} }
\end{array}\]where d will be equal to 2l because the circle cuts the y axis at the two points and d is the length between the two points. Hence, we will get a suitable answer.
Complete step by step solution:
In this question, we have given the circle \[\begin{array}{*{20}{c}}
{{x^2} + {y^2} + 2gx + 2fy + c}& = &0
\end{array}\]. We know that the center of the circle is C (-g, -f).
According to the given condition, we will write,
\[\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &{2\sqrt {{g^2} - c} }
\end{array}\]
In the above expression, d will be zero because the circle touches the x axis. Therefore, we can write
\[ \Rightarrow \begin{array}{*{20}{c}}
0& = &{2\sqrt {{g^2} - c} }
\end{array}\]
And we will get,
\[ \Rightarrow \begin{array}{*{20}{c}}
{{g^2}}& = &c
\end{array}\] ……………… (1).
And we have given that the circle cuts the y axis at the two points and distance between those points are also given, therefore, we will write,
\[\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &{2\sqrt {{f^2} - c} }
\end{array}\]
And we have,
\[\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &{2l}
\end{array}\]
Now put this value in the above expression, therefore, we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow 2l}& = &{2\sqrt {{f^2} - c} }
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow l}& = &{\sqrt {{f^2} - c} }
\end{array}\]
Now square both the sides,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {l^2}}& = &{{f^2} - c}
\end{array}\]
Now from the equation (1). We will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow {l^2}}& = &{{f^2} - {g^2}}
\end{array}\]
Now we know that the x = -g and y = -f . Therefore, the locus of the circle will be,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {l^2}}& = &{{y^2} - {x^2}}
\end{array}\]
Now the above equation is the equation of the locus of the hyperbola.
Therefore, the correct option is (D).
Note: The first point is to keep in mind that if the circle touches any axis then the cut-off length will be zero. But if the circle cuts off the circle at the axis, then the cut-off length will not be zero.
d& = &{2\sqrt {{g^2} - c} }
\end{array}\]where d will be zero because the circle touches the x axis. And then apply the formula \[\begin{array}{*{20}{c}}
d& = &{2\sqrt {{f^2} - c} }
\end{array}\]where d will be equal to 2l because the circle cuts the y axis at the two points and d is the length between the two points. Hence, we will get a suitable answer.
Complete step by step solution:
In this question, we have given the circle \[\begin{array}{*{20}{c}}
{{x^2} + {y^2} + 2gx + 2fy + c}& = &0
\end{array}\]. We know that the center of the circle is C (-g, -f).
According to the given condition, we will write,
\[\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &{2\sqrt {{g^2} - c} }
\end{array}\]
In the above expression, d will be zero because the circle touches the x axis. Therefore, we can write
\[ \Rightarrow \begin{array}{*{20}{c}}
0& = &{2\sqrt {{g^2} - c} }
\end{array}\]
And we will get,
\[ \Rightarrow \begin{array}{*{20}{c}}
{{g^2}}& = &c
\end{array}\] ……………… (1).
And we have given that the circle cuts the y axis at the two points and distance between those points are also given, therefore, we will write,
\[\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &{2\sqrt {{f^2} - c} }
\end{array}\]
And we have,
\[\begin{array}{*{20}{c}}
{ \Rightarrow d}& = &{2l}
\end{array}\]
Now put this value in the above expression, therefore, we will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow 2l}& = &{2\sqrt {{f^2} - c} }
\end{array}\]
\[\begin{array}{*{20}{c}}
{ \Rightarrow l}& = &{\sqrt {{f^2} - c} }
\end{array}\]
Now square both the sides,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {l^2}}& = &{{f^2} - c}
\end{array}\]
Now from the equation (1). We will get
\[\begin{array}{*{20}{c}}
{ \Rightarrow {l^2}}& = &{{f^2} - {g^2}}
\end{array}\]
Now we know that the x = -g and y = -f . Therefore, the locus of the circle will be,
\[\begin{array}{*{20}{c}}
{ \Rightarrow {l^2}}& = &{{y^2} - {x^2}}
\end{array}\]
Now the above equation is the equation of the locus of the hyperbola.
Therefore, the correct option is (D).
Note: The first point is to keep in mind that if the circle touches any axis then the cut-off length will be zero. But if the circle cuts off the circle at the axis, then the cut-off length will not be zero.
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