Question

The parametric equation of a circle passing through origin are given by \[ x=-g+5\cos\theta, \qquad y=-f+5\sin\theta. \] If the straight line passing through origin with slope \[ -\frac43 \] is the diameter of this circle, then the sum of the intercepts made by this circle on the coordinate axes is

• A. \(14\)
• B. \(7\)
• C. \(10\)
• D. \(11\)

Hint:

For \[ x=h+r\cos\theta,\qquad y=k+r\sin\theta, \] the centre is \((h,k)\) and the radius is \(r\). If the circle passes through the origin, \[ \boxed{h^2+k^2=r^2.} \]

Answer 1

The Correct Option is A

Step 1: Identify the centre and radius of the circle. The given parametric equations are \[ x=-g+5\cos\theta, \qquad y=-f+5\sin\theta. \] Hence, \[ \boxed{\text{Centre}=(-g,-f),\qquad r=5.} \] Since the circle passes through the origin, \[ g^2+f^2=25. \]

Step 2: Use the given diameter. The diameter passes through the origin and the centre. Hence the slope of the line joining \[ (0,0) \quad\text{and}\quad (-g,-f) \] is \[ \frac{-f}{-g}=\frac{f}{g}. \] Given slope \[ -\frac43, \] therefore, \[ \frac{f}{g}=-\frac43. \] Using \[ g^2+f^2=25, \] we obtain \[ g=3,\qquad f=-4. \] Thus the centre is \[ (-3,4). \]

Step 3: Find the intercepts. The equation of the circle is \[ (x+3)^2+(y-4)^2=25. \] For the \(x\)-axis, \[ y=0, \] giving \[ (x+3)^2=9, \] so the intercept length is \[ 6. \] For the \(y\)-axis, \[ x=0, \] giving \[ (y-4)^2=16, \] so the intercept length is \[ 8. \] Hence, \[ \boxed{6+8=14.} \] Therefore, the correct option is \(\boxed{(A)}\).