Question

A variable circle passes through the fixed point \(A(p,q)\) and touches the X-axis. The locus of the other end of the diameter through A is

• A. \((y-p)^2 = 4qx\)
• B. \((x-q)^2 = 4py\)
• C. \((x-p)^2 = 4qy\)
• D. \((y-q)^2 = 4px\)

Hint:

Whenever a circle touches a coordinate axis, replace the radius by the perpendicular distance of the center from that axis.

Answer 1

The Correct Option is C

Step 1: Name the moving endpoint.
Let the other diameter end be $P(x,y)$, with the fixed point $A(p,q)$. Since $AP$ is a diameter, the centre is their midpoint.
Step 2: Write the centre.
\[ C=\left(\frac{x+p}{2},\,\frac{y+q}{2}\right). \]
Step 3: Use the tangency to the X-axis.
Touching the X-axis means the radius equals the centre's height, so \[ R=\left|\frac{y+q}{2}\right|,\qquad R^2=\left(\frac{y+q}{2}\right)^2. \]
Step 4: Express $R$ as half the diameter.
The radius is also half of $AP$, so \[ R^2=\frac{(x-p)^2+(y-q)^2}{4}. \]
Step 5: Equate the two expressions.
Multiplying by $4$, \[ (y+q)^2=(x-p)^2+(y-q)^2. \]
Step 6: Expand and cancel.
The $y^2$ and $q^2$ terms cancel, leaving $2qy=(x-p)^2-2qy$, hence $(x-p)^2=4qy$. This is option 3.
\[ \boxed{(x-p)^2=4qy} \]