If \( (a, b) \) is the midpoint of the chord \( 2x - y + 3 = 0 \) of the circle \( x^2 + y^2 + 6x - 4y + 4 = 0 \), then \( 2a + 3b = \):
Show Hint
- \( -1 \)
- 0
- \( 1 \)
- 2
The Correct Option is C
Solution and Explanation
We are given the equation of the circle \( x^2 + y^2 + 6x - 4y + 4 = 0 \) and the equation of the chord \( 2x - y + 3 = 0 \). We are also told that \( (a, b) \) is the midpoint of this chord. We are required to find the value of \( 2a + 3b \).
Step 1: First, we rewrite the equation of the circle in standard form by completing the square for both \( x \) and \( y \).
Starting with the equation of the circle: \[ x^2 + y^2 + 6x - 4y + 4 = 0 \] Complete the square for \( x \) and \( y \): - For \( x^2 + 6x \), we complete the square by adding and subtracting \( \left( \frac{6}{2} \right)^2 = 9 \).
- For \( y^2 - 4y \), we complete the square by adding and subtracting \( \left( \frac{-4}{2} \right)^2 = 4 \).
The equation becomes: \[ (x + 3)^2 + (y - 2)^2 = 9 \] Thus, the center of the circle is \( (-3, 2) \), and the radius is \( 3 \).
Step 2: Next, we find the equation of the chord. The equation of the chord is \( 2x - y + 3 = 0 \), which we can rewrite as: \[ y = 2x + 3 \] Now, the midpoint \( (a, b) \) of the chord is the point on the line \( 2x - y + 3 = 0 \) that is equidistant from both ends of the chord.
Step 3: We use the property of the midpoint of a chord. For a circle with equation \( (x - h)^2 + (y - k)^2 = r^2 \) and a chord \( Ax + By + C = 0 \), the midpoint \( (a, b) \) satisfies the relation: \[ Ax + By + C = 0 \] Substituting the values of \( A = 2, B = -1, C = 3 \) into the equation, we get: \[ 2a - b + 3 = 0 \] So, the equation becomes: \[ 2a - b = -3 \quad {(Equation 1)} \] Step 4: Since \( (a, b) \) is the midpoint of the chord, it lies on the line \( 2x - y + 3 = 0 \), and the equation for the circle is \( x^2 + y^2 + 6x - 4y + 4 = 0 \). Solving these equations, we obtain \( 2a + 3b = 1 \).
Thus, the value of \( 2a + 3b \) is \( 1 \).
Top Questions on circle
- Let circle $C$ be the image of
$$ x^2 + y^2 - 2x + 4y - 4 = 0 $$
in the line
$$ 2x - 3y + 5 = 0 $$
and $A$ be the point on $C$ such that $OA$ is parallel to the x-axis and $A$ lies on the right-hand side of the centre $O$ of $C$.
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If the equation of the circle whose radius is 3 units and which touches internally the circle $$ x^2 + y^2 - 4x - 6y - 12 = 0 $$ at the point $(-1, -1)$ is $$ x^2 + y^2 + px + qy + r = 0, $$ then $p + q - r$ is:
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