Given that the circle having radius 4 units touches the coordinate axes in the first quadrant.
Let us assume the circle touches the co - ordinate axes at (a,0) and (0,a). Then the circle will have the centre at (a, a) and radius |a|.
It is given that the radius is 4 units. Since the circle touches the axes in the first quadrant, we will have the centre in the first quadrant.
The centre of the circle is (4, 4).
The centres of the mirrors of the circle w.r.t x = 0 and y = 0 is ( - 4, 4) and (4, - 4).
Case (i):
We have a circle with centre ( - 4, 4) and having radius 4.
We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:
⇒ (x - p)2 + (y - q)2 = r2
Now we substitute the corresponding values in the equation:
⇒ (x - ( - 4))2 + (y - 4)2 = 42
⇒ (x + 4)2 + (y - 4)2 = 16
⇒ x2 + 8x + 16 + y2 - 8y + 16 = 16
⇒ x2 + y2 + 8x - 8y + 16 = 0.
∴ The equation of the circle is x2 + y2 + 8x - 8y + 16 = 0.
Case (ii):
We have circle with centre (4, - 4) and having radius 4.
We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:
⇒ (x - p)2 + (y - q)2 = r2
Now we substitute the corresponding values in the equation:
⇒ (x - 4)2 + (y - ( - 4))2 = 42
⇒ (x - 4)2 + (y + 4)2 = 16
⇒ x2 - 8x + 16 + y2 + 8y + 16 = 16
⇒ x2 + y2 - 8x + 8y + 16 = 0.
∴ The equation of the circle is x2 + y2 - 8x + 8y + 16 = 0.
