VerifiedHint: Draw a circle in the first quadrant touch the coordinate axes at a distance of 1 unit from the origin at point A and point B. Now, get the coordinates of the points A and B. Since the axes are touching the circle so, the coordinate axes are tangent to the circle and the radius is perpendicular to the point of tangency. So, CA and CB are perpendicular to the x-axis and the y-axis respectively. Now, get the coordinate of the point C. Then, calculate the distance CA which is the radius of the circle by using distance formula, \[\text{Distance=}\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\] . We know the equation of a circle, \[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{\left( radius \right)}^{2}}\] . Here, \[\left( {{x}_{1}},{{y}_{1}} \right)\] is the coordinate of the center of the circle. In the given circle point C is the center of the circle. Now, put the coordinates of point C and the measure of the radius in the equation, \[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{\left( radius \right)}^{2}}\] . Solve it further and get the equation of the circle.
Complete step by step solution:
According to the question, it is given that a circle is in the first quadrant and it is touching each coordinate axis at a distance of one unit from the origin.
Let us draw a diagram for this.
We can observe that the circle is touching the x-axis at a distance of 1 unit from the origin at point A. As the given circle is touching the x-axis, so the point where the circle is touching has y coordinate equal to zero. The circle is also touching the x-axis at a distance of 1 unit. So, the point where the circle is touching has x-coordinate equal to 1.
Therefore, the coordinate of point A is (1,0) …………………………………..(1)
The circle is touching the y-axis at a distance of 1 unit from the origin at point B. As the given circle is touching the y-axis, so the point where the circle is touching has x coordinate equal to zero. The circle is also touching the y-axis at a distance of 1 unit. So, the point where the circle is touching has y-coordinate equal to 1.
Therefore, the coordinate of point B is (0,1) …………………………………(2)
It is given that the circle is touching the coordinate axes, so we can say that the coordinate axes are tangent to the given circle.
Now, joining C to A and C to B, we get the line CA and CB respectively. From the figure, we can see that the distance CA and CB is the radius of the given circle.
We know the property that the radius is perpendicular to the tangent of a circle. So, CA is perpendicular to the x-axis and CB is perpendicular to the y-axis.
Since the line, CA is perpendicular to the x-axis, so x-coordinate of point C is the same as of the x coordinate of point A.
Therefore, x- coordinate of point C is 1 ………………………….(3)
Since the line, CB is perpendicular to the y-axis, so the y-coordinate of point C is the same as the y-coordinate of point B.
Therefore, y- coordinate of point C is 1 ………………………………(4)
From equation (3) and equation (4), we have,
The coordinate of point C is (1,1) …………………………(5)
We have the distance CA as the radius of the circle.
Now, using the distance formula, we can get the distance CA.
From equation (1) and equation (5), we have,
The coordinate of point A = (1,0).
The coordinate of point C = (1,1).
$\text{Distance=}\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}} $
$\Rightarrow CA=\sqrt{{{\left( 1-1 \right)}^{2}}+{{\left( 1-0 \right)}^{2}}} $
$\Rightarrow CA=\sqrt{{{1}^{2}}} $
$ \Rightarrow CA=1 $
Therefore, the radius of the circle is equal to 1 …………………………………(6)
We know the equation of a circle, \[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{\left( radius \right)}^{2}}\] …………………………..(7)
Here, \[\left( {{x}_{1}},{{y}_{1}} \right)\] is the coordinate of the center of the circle.
In the given circle, point C is the center of the circle.
Now, from equation (5), equation (6), and equation (7), we get
${{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{\left( 1 \right)}^{2}} $
$\Rightarrow {{x}^{2}}-2x+1+{{y}^{2}}-2y+1=1 $
$\Rightarrow {{x}^{2}}+{{y}^{2}}-2x-2y+1=0 $
Therefore, the equation of the circle is \[{{x}^{2}}+{{y}^{2}}-2x-2y+1=0\] .
Hence, the correct option is (A).
Note: In this question, one might draw the figure of the circle as given below.
This figure is completely wrong because in question it is said that our circle lies in the first quadrant and this drawn circle is in the second quadrant. Therefore, we have to think of only that circle which is lying in the first quadrant. Thus, our required circle will look like below.
