
The equation of a circle which passes through the origin and cuts off intercepts 4 and 6 on the axes is
(1) ${x^2} + {y^2} = 24$
(2) ${x^2} + {y^2} - 4x - 6y = 0$
(3) ${x^2} + {y^2} = 10$
(4) ${x^2} + {y^2} - 8x - 12y = 0$
Answer
578.1k+ views
Hint:
Let the equation of the circle that passes through origin be ${x^2} + {y^2} + 2gx + 2fy = 0$. Now, according to the given condition, the circle will also pass through $\left( {4,0} \right)$ as \[x\] intercept is 4 and also passes through $\left( {0,6} \right)$ as $y$ intercept is of 6 units. Then, substitute these values to find the value of $f$ and $g$ and hence substitute them in the equation of the circle.
Complete step by step solution:
The standard equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$
But, we are given that the circle passes through origin, then the value of $c$ is zero.
Hence, the equation of the required circle is of the form ${x^2} + {y^2} + 2gx + 2fy = 0$
We are given that the circle cuts off the intercept on the $x$ axis of 4 units.
That is the point $\left( {4,0} \right)$ satisfies the equation of the circle.
On substituting the values of \[x = 4\] and $y = 0$ in the above equation.
$
{\left( 4 \right)^2} + {\left( 0 \right)^2} + 2g\left( 4 \right) + 2f\left( 0 \right) = 0 \\
\Rightarrow 16 + 8g = 0 \\
\Rightarrow 8g = - 16 \\
$
Divide the equation by 8
$g = - 2$
Similarly, we are given that the $y$ intercept is of 6 units. Therefore, the circle passes through the points $\left( {0,6} \right)$
That is the point $\left( {0,6} \right)$ satisfies the equation of the circle.
On substituting the values of \[x = 0\] and $y = 6$ in the above equation.
$
{\left( 0 \right)^2} + {\left( 6 \right)^2} + 2g\left( 0 \right) + 2f\left( 6 \right) = 0 \\
\Rightarrow 36 + 12f = 0 \\
\Rightarrow 12f = - 36 \\
$
Divide the equation by 12
$f = - 3$
Hence, the required equation of the circle is
$
\Rightarrow {x^2} + {y^2} + 2\left( { - 2} \right)x + 2\left( { - 3} \right)y = 0 \\
\Rightarrow {x^2} + {y^2} - 4x - 6y = 0 \\
$
Thus, option (2) is correct.
Note:
We can also use the equation of circle, ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, where $\left( {h,k} \right)$ are the coordinates of the centre and $r$ is the radius of the circle. We know that the circle passes through $\left( {0,0} \right)$, $\left( {4,0} \right)$ and $\left( {0,6} \right)$. We can form three equations with these points and solve them simultaneously to find the unknown values.
Let the equation of the circle that passes through origin be ${x^2} + {y^2} + 2gx + 2fy = 0$. Now, according to the given condition, the circle will also pass through $\left( {4,0} \right)$ as \[x\] intercept is 4 and also passes through $\left( {0,6} \right)$ as $y$ intercept is of 6 units. Then, substitute these values to find the value of $f$ and $g$ and hence substitute them in the equation of the circle.
Complete step by step solution:
The standard equation of a circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$
But, we are given that the circle passes through origin, then the value of $c$ is zero.
Hence, the equation of the required circle is of the form ${x^2} + {y^2} + 2gx + 2fy = 0$
We are given that the circle cuts off the intercept on the $x$ axis of 4 units.
That is the point $\left( {4,0} \right)$ satisfies the equation of the circle.
On substituting the values of \[x = 4\] and $y = 0$ in the above equation.
$
{\left( 4 \right)^2} + {\left( 0 \right)^2} + 2g\left( 4 \right) + 2f\left( 0 \right) = 0 \\
\Rightarrow 16 + 8g = 0 \\
\Rightarrow 8g = - 16 \\
$
Divide the equation by 8
$g = - 2$
Similarly, we are given that the $y$ intercept is of 6 units. Therefore, the circle passes through the points $\left( {0,6} \right)$
That is the point $\left( {0,6} \right)$ satisfies the equation of the circle.
On substituting the values of \[x = 0\] and $y = 6$ in the above equation.
$
{\left( 0 \right)^2} + {\left( 6 \right)^2} + 2g\left( 0 \right) + 2f\left( 6 \right) = 0 \\
\Rightarrow 36 + 12f = 0 \\
\Rightarrow 12f = - 36 \\
$
Divide the equation by 12
$f = - 3$
Hence, the required equation of the circle is
$
\Rightarrow {x^2} + {y^2} + 2\left( { - 2} \right)x + 2\left( { - 3} \right)y = 0 \\
\Rightarrow {x^2} + {y^2} - 4x - 6y = 0 \\
$
Thus, option (2) is correct.
Note:
We can also use the equation of circle, ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, where $\left( {h,k} \right)$ are the coordinates of the centre and $r$ is the radius of the circle. We know that the circle passes through $\left( {0,0} \right)$, $\left( {4,0} \right)$ and $\left( {0,6} \right)$. We can form three equations with these points and solve them simultaneously to find the unknown values.
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