Question

The area of the quadrilateral formed by the common tangents drawn to the circle \[ x^2+y^2=16 \] and the ellipse \[ 7x^2+25y^2=175 \] is:

• A. \(64\)
• B. \(32\)
• C. \(5\sqrt2\)
• D. \(16\sqrt2\)

Hint:

For a tangent \(y=mx+c\) to the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] remember the condition \[ c^2=a^2m^2+b^2. \] Equating this with the tangency condition of another conic is often the fastest way to obtain common tangents.

Answer 1

The Correct Option is A

Concept: A line will be a common tangent to both the circle and the ellipse if it satisfies the tangency condition for each curve simultaneously. The circle \[ x^2+y^2=16 \] has centre at the origin and radius \(4\). The ellipse \[ 7x^2+25y^2=175 \] can be written as \[ \frac{x^2}{25}+\frac{y^2}{7}=1. \] The common tangents obtained will form a quadrilateral. After determining the equations of these tangents, we find the vertices of the quadrilateral and then compute its area.

Step 1: Write the ellipse in standard form. Dividing by \(175\), \[ \frac{x^2}{25}+\frac{y^2}{7}=1. \] Thus, \[ a^2=25,\qquad b^2=7. \] Hence \[ a=5,\qquad b=\sqrt7. \]

Step 2: Take the equation of a tangent in slope form. Let the tangent be \[ y=mx+c. \] For the ellipse \[ \frac{x^2}{25}+\frac{y^2}{7}=1, \] the condition of tangency is \[ c^2=25m^2+7. \]

Step 3: Use the tangency condition for the circle. For the circle \[ x^2+y^2=16, \] the distance of the centre \((0,0)\) from the tangent must equal the radius \(4\). Therefore, \[ \frac{|c|}{\sqrt{1+m^2}}=4. \] Squaring, \[ c^2=16(1+m^2). \]

Step 4: Equate the two values of \(c^2\). Since the line is tangent to both curves, \[ 25m^2+7=16(1+m^2). \] Expanding, \[ 25m^2+7=16+16m^2. \] \[ 9m^2=9. \] \[ m^2=1. \] \[ m=\pm1. \]

Step 5: Find the corresponding values of \(c\). Using \[ c^2=16(1+m^2), \] and \(m^2=1\), \[ c^2=16(2)=32. \] \[ c=\pm4\sqrt2. \] Hence the four common tangents are \[ y=x+4\sqrt2, \] \[ y=x-4\sqrt2, \] \[ y=-x+4\sqrt2, \] \[ y=-x-4\sqrt2. \]

Step 6: Determine the vertices of the quadrilateral. Intersecting \[ y=x+4\sqrt2 \] and \[ y=-x+4\sqrt2, \] gives \[ x=0,\qquad y=4\sqrt2. \] Similarly, the four vertices are \[ (0,4\sqrt2), \] \[ (4\sqrt2,0), \] \[ (0,-4\sqrt2), \] \[ (-4\sqrt2,0). \] Thus the quadrilateral is a square. Its diagonals are \[ 8\sqrt2 \] and \[ 8\sqrt2. \]

Step 7: Compute the area. Area of a rhombus (or square) in terms of diagonals: \[ \text{Area} = \frac12 d_1d_2. \] Hence \[ \text{Area} = \frac12(8\sqrt2)(8\sqrt2). \] \[ = \frac12(128). \] \[ =64. \]

Step 8: Final Answer. \[ \boxed{64} \]