Question

A line meets the circle $x^2+y^2-4x-4y-8=0$ in two points A and B. If P(2,-2) is a point on the circle such that PA = PB = 2 then the equation of the line AB is

• A. $2x+3y=0$
• B. $3x+2y=0$
• C. $2x+3=0$
• D. $2y+3=0$

Hint:

When a point P on a circle is equidistant from the endpoints of a chord AB, the line joining the center C and P is the perpendicular bisector of the chord AB. This geometric property is often the key to solving such problems efficiently.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The line AB is a chord of the given circle. P is a point on the circle such that it is equidistant from A and B ($PA=PB$). This means that P lies on the perpendicular bisector of the chord AB. Since the perpendicular bisector of any chord passes through the center of the circle, the line joining the center (C) and the point P must be perpendicular to the chord AB.
Step 2: Key Formula or Approach
1. Find the center C of the given circle. 2. Find the slope of the line segment CP. 3. The line AB is perpendicular to CP. Find the slope of AB. 4. We also need a point on the line AB. Let M be the midpoint of AB. The points C, M, and P are collinear. We can find the distance CM to locate M. 5. Use the point-slope form to find the equation of line AB.
Step 3: Detailed Explanation
1. Find the center and radius of the circle: The equation is $x^2+y^2-4x-4y-8=0$. Center $C = (-g, -f) = (-(-2), -(-2)) = (2,2)$. Radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-2)^2+(-2)^2-(-8)} = \sqrt{4+4+8} = \sqrt{16}=4$. 2. Find the slope of CP: The points are $C(2,2)$ and $P(2,-2)$. The slope of CP is $m_{CP} = \frac{-2-2}{2-2} = \frac{-4}{0}$, which is undefined. This means CP is a vertical line. The equation of the line CP is $x=2$. 3. Find the slope and equation type of AB: Since the chord AB is perpendicular to the vertical line CP, AB must be a horizontal line. The slope of AB is 0, and its equation is of the form $y=k$ for some constant $k$. This constant $k$ is the y-coordinate of the midpoint M of AB. 4. Find the location of M: Triangle CMA is a right-angled triangle with hypotenuse CA (the radius). $CA=r=4$. Triangle PMA is a right-angled triangle with hypotenuse PA. We are given $PA=2$. Let $AM$ be half the length of the chord. In $\triangle CMA$, $CM^2 + AM^2 = CA^2 = 4^2 = 16$. In $\triangle PMA$, $PM^2 + AM^2 = PA^2 = 2^2 = 4$. The points C, M, P are collinear. The distance $CP = \sqrt{(2-2)^2 + (2-(-2))^2} = \sqrt{0+16} = 4$. Since M lies on the segment CP, $CP = CM+PM = 4$. From the two triangle equations, we have $AM^2 = 16-CM^2$ and $AM^2 = 4-PM^2$. So, $16-CM^2 = 4-PM^2$. Also $PM=4-CM$. $16-CM^2 = 4-(4-CM)^2 = 4-(16-8CM+CM^2) = -12+8CM-CM^2$. $16 = -12+8CM \implies 28 = 8CM \implies CM = \frac{28}{8} = \frac{7}{2}$. The point M lies on the vertical line segment CP ($x=2$) at a distance of $7/2$ from C(2,2). Since P(2,-2) is 'below' C(2,2), M must also be below C. The coordinates of M are $(2, 2-CM) = (2, 2-7/2) = (2, -3/2)$. The y-coordinate of M is $-3/2$. 5. Equation of line AB: Since AB is a horizontal line passing through M(2, -3/2), its equation is $y = -3/2$. Rearranging gives $2y = -3$, or $2y+3=0$. Step 4: Final Answer
The equation of the line AB is $2y+3=0$.