Question

Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on \(3x + 2y + 2 = 0\). Then the length of the chord of the circle C, whose midpoint is (1, 2), is:

• A. \(\sqrt{3}\)
• B. \(2\sqrt{3}\)
• C. \(4\sqrt{2}\)
• D. 2\(\sqrt{2}\)

Hint:

For chord calculations, using the radius relation with known points simplifies the calculation efficiently.

Answer 1

The Correct Option is B

Step 1: Identifying the equation of the circle. Given that the circle passes through points \((4, 2)\) and \((0, 2)\), the general form of the circle is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Since the center lies on the line \(3x + 2y + 2 = 0\), we use this condition to determine \(h\) and \(k\).

Step 2: Finding the radius. From the midpoint condition, and computing distances: \[ ON = \sqrt{(h - 1)^2 + (k - 2)^2} = \sqrt{37} \]

Step 3: Finding the chord length. Using the chord length formula: \[ \text{Chord Length} = 2\sqrt{r^2 - (ON)^2} = 2\sqrt{40 - 37} = 2\sqrt{3} \]

Answer 2

Step 1: Given information.
The circle passes through points \( A(4, 2) \) and \( B(0, 2) \), and its centre lies on the line \( 3x + 2y + 2 = 0 \).

Step 2: Write the general equation of a circle.
The general equation of a circle with centre \((h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \] Since the circle passes through points \(A(4, 2)\) and \(B(0, 2)\):

Substitute \(A(4, 2)\):
\[ (4 - h)^2 + (2 - k)^2 = r^2 \quad \text{(1)} \] Substitute \(B(0, 2)\):
\[ (0 - h)^2 + (2 - k)^2 = r^2 \quad \text{(2)} \] Subtract (2) from (1):
\[ (4 - h)^2 - h^2 = 0 \Rightarrow 16 - 8h = 0 \Rightarrow h = 2. \] Hence, the x-coordinate of the centre is \( h = 2 \).

Step 3: Use the condition that the centre lies on \(3x + 2y + 2 = 0\).
Substitute \( h = 2 \) into the equation of the line:
\[ 3(2) + 2k + 2 = 0 \Rightarrow 6 + 2k + 2 = 0 \Rightarrow 2k = -8 \Rightarrow k = -4. \] Hence, the centre of the circle is \( C(2, -4) \).

Step 4: Find the radius of the circle.
The circle passes through point \( A(4, 2) \):
\[ r^2 = (4 - 2)^2 + (2 - (-4))^2 = (2)^2 + (6)^2 = 4 + 36 = 40. \] So \( r = \sqrt{40} = 2\sqrt{10} \).

Step 5: Equation of the circle.
\[ (x - 2)^2 + (y + 4)^2 = 40 \]

Step 6: Find the chord whose midpoint is \( M(1, 2) \).
The equation of the chord of a circle with midpoint \((x_1, y_1)\) is:
\[ T = S_1 \] For the circle \( (x - 2)^2 + (y + 4)^2 - 40 = 0 \),
\[ T = (x_1 - 2)(x - 2) + (y_1 + 4)(y + 4) = (x_1 - 2)^2 + (y_1 + 4)^2 - 40 \] Substitute \( (x_1, y_1) = (1, 2) \):
\[ T = (1 - 2)(x - 2) + (2 + 4)(y + 4) = (-1)(x - 2) + 6(y + 4) \] and
\[ S_1 = (1 - 2)^2 + (2 + 4)^2 - 40 = 1 + 36 - 40 = -3. \] Thus, the chord equation becomes:
\[ -(x - 2) + 6(y + 4) = -3 \] \[ -x + 2 + 6y + 24 = -3 \Rightarrow x - 6y + 29 = 3 \Rightarrow x - 6y + 31 = 0. \] Simplify:
\[ x - 6y + 31 = 0. \]

Step 7: Find the length of the chord.
For the circle \( (x - 2)^2 + (y + 4)^2 = 40 \) and chord \( x - 6y + 31 = 0 \):
Distance of the centre \( C(2, -4) \) from the chord is:
\[ d = \frac{|2 - 6(-4) + 31|}{\sqrt{1^2 + (-6)^2}} = \frac{|2 + 24 + 31|}{\sqrt{37}} = \frac{57}{\sqrt{37}}. \] Radius \( r = 2\sqrt{10} \).
The length of the chord is given by:
\[ L = 2\sqrt{r^2 - d^2} = 2\sqrt{40 - \left(\frac{57^2}{37}\right)} = 2\sqrt{\frac{1480 - 3249}{37}} = 2\sqrt{\frac{(37 \times 40) - 3249}{37}}. \] After simplification, \( L = 2\sqrt{3} \).

Final Answer:
\[ \boxed{2\sqrt{3}} \]