Question

Let \(C\) be a circle having centre in the first quadrant and touching the \(x\)-axis at a distance of \(3\) units from the origin. If the circle \(C\) has an intercept of length \(6\sqrt{3}\) on \(y\)-axis, then the length of the chord of the circle \(C\) on the line \(x-y=3\) is:

• A. \(8\)
• B. \(6\)
• C. \(6\sqrt{2}\)
• D. \(8\sqrt{2}\)

Answer 1

The Correct Option is C

Concept: If a line cuts a chord in a circle, the chord length can be computed using: \[ \text{Chord length} = 2\sqrt{r^2-d^2} \] where \(d\) is the perpendicular distance from the center to the line.

Step 1: {Determine the centre and radius of the circle.} Since the circle touches the \(x\)-axis at \((3,0)\), the centre lies vertically above this point. Let centre be: \[ (3,r) \] Radius \(=r\). 

Step 2: {Use the intercept on the \(y\)-axis.} Distance from centre \((3,r)\) to the \(y\)-axis: \[ =3 \] Length of intercept on the \(y\)-axis: \[ 2\sqrt{r^2-3^2} \] Given: \[ 2\sqrt{r^2-9}=6\sqrt{3} \] \[ \sqrt{r^2-9}=3\sqrt{3} \] \[ r^2-9=27 \] \[ r^2=36 \] \[ r=6 \] Thus centre: \[ (3,6) \] 

Step 3: {Find perpendicular distance from centre to the line \(x-y=3\).} Line form: \[ x-y-3=0 \] Distance: \[ d=\frac{|3-6-3|}{\sqrt{1^2+(-1)^2}} \] \[ =\frac{6}{\sqrt2}=3\sqrt2 \] 

Step 4: {Find the chord length.} \[ \text{Chord length}=2\sqrt{36-(3\sqrt2)^2} \] \[ =2\sqrt{36-18} \] \[ =2\sqrt{18} \] \[ =6\sqrt2 \]