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: If a line \[ Ax+By+C=0 \] is tangent to a circle with centre \((h,k)\) and radius \(r\), then \[ \frac{|Ah+Bk+C|} {\sqrt{A^2+B^2}} = r. \]

Step 1: Find the tangent to the circle \(x^2+y^2=3\) at \((\sqrt2,1)\). For the circle \[ x^2+y^2=3, \] the tangent at \((x_1,y_1)\) is \[ xx_1+yy_1=3. \] At \[ (\sqrt2,1), \] the tangent is \[ \sqrt2\,x+y-3=0. \]

Step 2: Use the tangency condition for the circles with centres \((0,\beta)\). The radius of each circle is \[ 2\sqrt3. \] Distance from \((0,\beta)\) to the line \[ \sqrt2\,x+y-3=0 \] must be \(2\sqrt3\). Hence, \[ \frac{|\,\beta-3\,|} {\sqrt{(\sqrt2)^2+1^2}} = 2\sqrt3. \] \[ \frac{|\,\beta-3\,|}{\sqrt3} = 2\sqrt3. \] \[ |\,\beta-3\,| = 6. \]

Step 3: Find the two possible values of \(\beta\). \[ \beta-3=6 \] or \[ \beta-3=-6. \] Thus, \[ \beta_1=9, \qquad \beta_2=-3. \]

Step 4: Compute \(|\beta_1-\beta_2|\). \[ |\beta_1-\beta_2| = |9-(-3)|. \] \[ =12. \]

Step 5: Write the final answer. \[ \boxed{12} \]