Question

Let the centre of a circle, passing through the point \((0, 0)\), \((1, 0)\) and touching the circle \(x^2 + y^2 = 9\), be \((h, k)\). Then for all possible values of the coordinates of the centre \((h, k)\), \(4(h^2 + k^2)\) is equal to __________.

Answer 1

Correct Answer: 9

Step 1: General equation of the circle The equation of the circle is:

\[ (x - h)^2 + (y - k)^2 = r^2. \]

Since the circle passes through the points \((0, 0)\) and \((1, 0)\), substitute these points into the circle’s equation.

For \((0, 0)\):

\[ h^2 + k^2 = r^2. \]

For \((1, 0)\):

\[ (1 - h)^2 + k^2 = r^2. \]

Expanding and simplifying:

\[ 1 - 2h + h^2 + k^2 = h^2 + k^2. \]

Substitute \(r^2 = h^2 + k^2\) into the equation:

\[ 1 - 2h + h^2 + k^2 = h^2 + k^2. \]

Cancel \(h^2 + k^2\):

\[ 1 - 2h = 0 \implies h = \frac{1}{2}. \]

Step 2: Circle touches \(x^2 + y^2 = 9\) The given circle \(x^2 + y^2 = 9\) has a radius \(R = 3\) and is centered at \((0, 0)\). For the circle to touch \(x^2 + y^2 = 9\), the distance between their centers must be equal to the difference of their radii:

\[ \sqrt{h^2 + k^2} = R - r = 3 - \sqrt{h^2 + k^2}. \]

Let \(d = \sqrt{h^2 + k^2}\):

\[ d = 3 - d \implies 2d = 3 \implies d = \frac{3}{2}. \]

Thus:

\[ h^2 + k^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}. \]

Step 3: Compute \(4(h^2 + k^2)\) Multiply \(h^2 + k^2\) by 4:

\[ 4(h^2 + k^2) = 4 \cdot \frac{9}{4} = 9. \]

Final Answer: 9.