Solution:
Let's find the equation of the circle which satisfies the given conditions:
The circle touches the x-axis.
The circle passes through the point (1,1).
The centre of the circle lies on the line x+y=3.
Let the centre of the circle be (h,k). Since the circle touches the x-axis, the distance from the centre to the x-axis (which is k ) is equal to the radius r of the circle. Thus, we have:
r=k
Next, since the line x+y=3 is known to contain the centre (h,k), we can write:
h+k=3
Now, the circle also passes through the point (1,1). Using the distance formula, the distance from the centre (h,k) to this point should also equal the radius r, so we can write:
√(h−1)2+(k−1)2=r
Since r=k, we substitute this in:
√(h−1)2+(k−1)2=k
Squaring both sides, we get:
(h−1)2+(k−1)2=k2
Expanding and simplifying:
h2−2h+1+k2−2k+1=k2
This simplifies further:
h2−2h+2=2k
Remember, from h+k=3, we can express k as:
k=3−h
Substitute k=3−h back into the equation:
h2−2h+2=2(3−h)
Simplify:
‌h2−2h+2=6−2h
‌h2+2=6
‌h2=4
So, we have:
h=2‌ or ‌h=−2
Since the centre lies in the first quadrant, h must be positive. Thus:
h=2
And, from h+k=3 :
2+k=3
Therefore:
k=1
Thus, the centre of the circle is (2,1) and its radius is 1 . The standard equation of the circle is:
(x−h)2+(y−k)2=r2
Substituting for (h,k) and r :
(x−2)2+(y−1)2=1
Expanding this equation, we have:
x2−4x+4+y2−2y+1=1
Combine like terms:
x2+y2−4x−2y+4=0
Thus, the equation of the circle is:
x2+y2−4x−2y+4=0
The correct option is:
Option B: x2+y2−4x−2y+4=0
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