Question

If the tangent drawn at \((\sqrt2,1)\) on the circle \[ x^2+y^2=3 \] is also a tangent to the two circles of equal radius \(2\sqrt3\) with centres at \[ (0,\beta_1) \quad \text{and} \quad (0,\beta_2), \] then \[ |\beta_1-\beta_2|= \]

• A. \(6\)
• B. \(12\)
• C. \(18\)
• D. \(5\) \bigskip

Hint:

For a circle tangent to a line, the perpendicular distance from the centre to the line is exactly equal to the radius: \[ \text{Distance}=\text{Radius}. \] This often gives two possible centre locations, one on each side of the tangent.

Answer 1

The Correct Option is B

Concept: The tangent line to \(x^2+y^2=3\) at a point is found using \(xx_1+yy_1=3\). Then impose the tangency condition for the second circle using the perpendicular distance formula from its centre to that line, equating to its radius.

Step 1: Tangent at \((\sqrt2,1)\): \(\sqrt2 x + y = 3\) or \(\sqrt2 x + y - 3 = 0\).

Step 2: Circle with centre \((0,\beta)\) and radius \(2\sqrt3\) is tangent to this line: \(\frac{|\sqrt2(0) + \beta - 3|}{\sqrt{(\sqrt2)^2+1^2}} = 2\sqrt3 \Rightarrow \frac{|\beta-3|}{\sqrt3} = 2\sqrt3 \Rightarrow |\beta-3| = 6\).

Step 3: Solve: \(\beta = 9\) or \(\beta = -3\). The difference \(|9 - (-3)| = 12\).

Step 4: Write the final answer. \(\boxed{12}\)