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

Concept: Let the other end of the diameter through the fixed point \(A(p,q)\) be \(P(x,y)\). Since \(A\) and \(P\) are the endpoints of a diameter, the center of the circle is the midpoint of \(AP\). Also, the circle touches the X-axis. Therefore, the radius is equal to the perpendicular distance of the center from the X-axis. The locus can be obtained by equating two expressions for the radius.

Step 1: Find the coordinates of the center of the circle. Since \(A(p,q)\) and \(P(x,y)\) are endpoints of a diameter, \[ \text{Center } C = \left( \frac{x+p}{2}, \frac{y+q}{2} \right) \] Because the circle touches the X-axis, \[ R=\left|\frac{y+q}{2}\right| \] Hence \[ R^2=\left(\frac{y+q}{2}\right)^2 \]

Step 2: Express the radius using the distance formula. The radius is also equal to the distance from the center to point \(A(p,q)\). Therefore, \[ R^2 = \left(\frac{x+p}{2}-p\right)^2 + \left(\frac{y+q}{2}-q\right)^2 \] \[ = \left(\frac{x-p}{2}\right)^2 + \left(\frac{y-q}{2}\right)^2 \] Thus, \[ R^2 = \frac{(x-p)^2+(y-q)^2}{4} \]

Step 3: Equate the two expressions for the radius. \[ \left(\frac{y+q}{2}\right)^2 = \frac{(x-p)^2+(y-q)^2}{4} \] Multiplying throughout by \(4\), \[ (y+q)^2 = (x-p)^2+(y-q)^2 \] Expanding, \[ y^2+2qy+q^2 = (x-p)^2+y^2-2qy+q^2 \] Cancelling common terms, \[ 2qy=(x-p)^2-2qy \] \[ (x-p)^2=4qy \] \[ \boxed{(x-p)^2=4qy} \]