A variable circle passes through the fixed point A( p,q) and touches x...
Let the variable circle be x2 + y2 + 2gx + 2fy + c = 0 ....(1)
∴ p2 + q2 + 2gp + 2fq + c =0 ....(2)
Circle (1) touches x-axis,
∴ g2 - c = 0 ⇒ c=g2 .
From (2) p2 + q2 + 2gp + 2fq +g2 =0 ....(3)
Let the other end of diameter through (p, q) be (h, k),
then

Put in (3)
⇒ h2 + p2 - 2hp - 4kq=0
∴ locus of (h, k) is x2 + p2 - 2 xp - 4 yq=0
⇒ ( x - p )2=4qy
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A variable circle passes through the fixed point A( p,q) and touches x...
Problem Analysis:
We are given a variable circle that passes through a fixed point A(p, q) and touches the x-axis. We need to determine the locus of the other end of the diameter through A.
Solution:
Let B(x, y) be the other end of the diameter through A.
Step 1: Find the equation of the circle:
Since the circle passes through A(p, q), the equation of the circle is given by:
(x - p)^2 + (y - q)^2 = r^2
where r is the radius of the circle.
Step 2: Find the equation of the tangent:
Since the circle touches the x-axis, the tangent at the point of contact will be parallel to the y-axis. Therefore, the equation of the tangent is given by:
x = p
Step 3: Find the coordinates of the point of contact:
Substituting x = p in the equation of the circle, we get:
(p - p)^2 + (y - q)^2 = r^2
0 + (y - q)^2 = r^2
(y - q)^2 = r^2
Therefore, the point of contact is (p, q).
Step 4: Find the coordinates of the other end of the diameter:
Since the diameter passes through the center of the circle, the coordinates of the center are (p, q). Therefore, the other end of the diameter is the reflection of A(p, q) with respect to the center (p, q). This can be obtained by using the formula for reflection:
x' = 2p - x
y' = 2q - y
Therefore, the coordinates of B(x, y) are:
x = 2p - p = p
y = 2q - q = q
Step 5: Find the equation of the locus:
Substituting x = p and y = q in the equation of the circle, we get:
(x - p)^2 + (y - q)^2 = r^2
(p - p)^2 + (q - q)^2 = r^2
0 + 0 = r^2
0 = r^2
Therefore, the equation simplifies to:
(x - p)^2 + (y - q)^2 = 0
Simplifying further, we get:
(x - p)^2 = 0
Therefore, the locus of the other end of the diameter through A is:
(x - p)^2 = 0
which can be written as:
(x - p)^2 = 4qy
Hence, the correct answer is option D: (x - p)^2 = 4qy.