Question

Find that the circle \[{x^2} + {y^2} - 8x + 4y + 4 = 0\] touches which of the axis?
A. x-axis
B. y-axis.
C. both axis
D. neither x-axis nor y-axis.

Answer 1

Hint: In this question, we start by comparing the given equation of the circle with the standard equation of the circle, from which we will find out the centre as well as radius of the circle given. After this, we will simply draw the diagram and find out the required answer.

Formula used:
The general equation of a circle is
\[{x^2} + {y^2} + 2gx + 2fy + c = 0\]
Here, the centre is \[( - g, - f)\] and the radius is \[\sqrt {{g^2} + {f^2} - c} \] .

Complete step by step solution:
The given equation is \[{x^2} + {y^2} - 8x + 4y + 4 = 0\].
Compare the given equation with the general equation to obtain the values of g, f and c.
Therefore, \[g = - 4,f = 2,c = 4\] .
Now, substitute the values \[g = - 4,f = 2,c = 4\] in the expression \[\sqrt {{g^2} + {f^2} - c} \]to obtain the radius.
Hence,
\[\sqrt {{{\left( { - 4} \right)}^2} + {2^2} - 4} \]
\[ = \sqrt {16 + 4 - 4} \]
\[ = 4\]
Therefore, the centre is (-4, 2) and radius is 4.
The distance of a point from x-axis is the ordinate of the point.
The distance of a point from y-axis is the abscissa of the point.
The distance of the point (-4,2) from the x-axis is 2 units.
 The distance of the point (-4,2) from the y-axis is -4 units.
But distance can’t be negative. Thus The distance of the point (-4,2) from the y-axis is 4 units.
Since the distance of the centre from y axis is same as the radius of the circle.
Thus the circle touches y-axis.
The correct option is B.

Note: We are given with the equation of the circle and the centre is at a point other than origin (0,0). Comparing the given equation of the circle with the general form of the equation, we can check where it touches the axis. Also, the condition for the circle to touch the x-axis, y must be zero and to touch y-axis, x must be equal to zero at touching point.