Question

If the length of the chord of the line \[ x+y-1=0 \] of the circle \[ x^2+y^2-6x+2fy-2=0\qquad(f>0) \] is \[ \frac{6\sqrt7}{\sqrt2}, \] then the length of the intercept made by this circle on the \(Y\)-axis is

• A. \(3\sqrt3\)
• B. \(12\sqrt3\)
• C. \(6\sqrt3\)
• D. \(4\sqrt3\)

Hint:

The length of a chord at perpendicular distance \(d\) from the centre is \[ \boxed{2\sqrt{r^2-d^2}}. \] To find the intercept on an axis, substitute the corresponding coordinate (\(x=0\) or \(y=0\)) in the circle equation.

Answer 1

The Correct Option is C

Step 1: Find the centre and radius of the circle. The given circle is \[ x^2+y^2-6x+2fy-2=0. \] Comparing with \[ x^2+y^2+2gx+2fy+c=0, \] we get \[ g=-3,\qquad c=-2. \] Hence, \[ \boxed{\text{Centre }=(3,-f)} \] and \[ r^2=g^2+f^2-c =9+f^2+2 =f^2+11. \]

Step 2: Use the chord length formula. The perpendicular distance of the centre from \[ x+y-1=0 \] is \[ d=\frac{|3-f-1|}{\sqrt2} =\frac{|2-f|}{\sqrt2}. \] The length of the chord is \[ 2\sqrt{r^2-d^2}. \] Given, \[ 2\sqrt{r^2-d^2} =\frac{6\sqrt7}{\sqrt2} =3\sqrt{14}. \] Squaring, \[ 4(r^2-d^2)=126, \] or \[ r^2-d^2=\frac{63}{2}. \] Substituting \[ r^2=f^2+11, \qquad d^2=\frac{(2-f)^2}{2}, \] we obtain \[ f^2+11-\frac{(2-f)^2}{2} =\frac{63}{2}. \] Multiplying by \(2\), \[ 2f^2+22-(f^2-4f+4)=63, \] \[ f^2+4f-45=0. \] Thus, \[ (f-5)(f+9)=0. \] Since \(f>0\), \[ \boxed{f=5.} \]

Step 3: Find the intercept on the \(Y\)-axis. Putting \[ x=0 \] in the circle, \[ y^2+10y-2=0. \] The intercept on the \(Y\)-axis equals the distance between the two points of intersection, \[ \sqrt{(10)^2-4(1)(-2)} = \sqrt{108} = 6\sqrt3. \] Therefore, \[ \boxed{6\sqrt3}. \] Hence, the correct option is \(\boxed{(C)}\).