Question

From a point A(0,3) on the circle \[ (x + 2)^2 + (y - 3)^2 = 4 \] a chord AB is drawn and extended to a point Q such that AQ = 2AB. Then the locus of Q is:

• A. \( (x + 4)^2 + (y - 3)^2 = 16 \)
• B. \( (x + 1)^2 + (y - 3)^2 = 32 \)
• C. \( (x + 1)^2 + (y - 3)^2 = 4 \)
• D. \( (x + 1)^2 + (y - 3)^2 = 1 \)

Hint:

Using the midpoint formula correctly ensures accurate derivation of the locus equation.

Answer 1

The Correct Option is A

The given equation of the circle is: \[ (x + 2)^2 + (y - 3)^2 = 4 \] Let the coordinates of \( Q(h,k) \). Since \( AQ = 2AB \), the midpoint \( B \) of segment \( AQ \) satisfies: \[ B = \left( \frac{0 + h}{2}, \frac{3 + k}{2} \right) \] Since point \( B \) lies on the given circle: \[ \left( \frac{h}{2} + 2 \right)^2 + \left( \frac{k}{2} - 3 \right)^2 = 4 \] Expanding: \[ \left( \frac{h + 4}{2} \right)^2 + \left( \frac{k - 3}{2} \right)^2 = 4 \] Multiplying both sides by 4: \[ (h + 4)^2 + (k - 3)^2 = 16 \] Thus, the required locus of \( Q(h,k) \) is: \[ (x + 4)^2 + (y - 3)^2 = 16 \]

Answer 2

From a point A(0,3) on the circle
\[ (x + 2)^2 + (y - 3)^2 = 4 \] a chord AB is drawn and extended to a point Q such that AQ = 2AB. Then the locus of Q is:
The center of the given circle is \( (-2, 3) \) and its radius is \( 2 \). Point \( A(0, 3) \) lies on this circle.
Let the coordinates of \( B(x_1, y_1) \) be on the circle. The midpoint of chord \( AB \) lies on the line joining the center of the circle and \( A \), which is along the x-axis.
Given that AQ = 2AB, point Q lies on a circle centered at \( (-4, 3) \) with a radius that is twice the radius of the original circle.
Thus, the locus of point Q is described by the equation:
\[ (x + 4)^2 + (y - 3)^2 = 16 \]
Therefore, the correct answer is:

\( (x + 4)^2 + (y - 3)^2 = 16 \)