if two distinct chords drawn from the point
(a,b) on the circle x2+y2=ax+by
when ab=0 are bisected by the x-axis
prove that a2=8b2
help me please i don,t know it
To prove that a² = 8b², we need to use the given information and properties of chords of a circle.
Given:
Equation of the circle: x² + y² = ax + by
Point on the circle: (a, b)
Condition: Two distinct chords drawn from this point, when ab = 0, are bisected by the x-axis.
Let's start by finding the coordinates of the points where the chords intersect the circle.
Using the equation of the circle, we substitute y = 0 to find the x-coordinate when the chord is bisected by the x-axis.
x² + 0² = ax + by
x² = ax ⇒ x = a (since b = 0)
So, the first point of intersection is (a, 0).
Similarly, substitute x = 0 in the equation of the circle to find the y-coordinate when the chord is bisected by the x-axis.
0² + y² = ax + by
y² = by ⇒ y = b (since a = 0)
Therefore, the second point of intersection is (0, b).
Now, let's find the midpoint M of the chord. The midpoint lies on the x-axis since the chord is bisected by the x-axis.
Midpoint coordinates:
M = ((a + 0) / 2, (0 + b) / 2)
M = (a/2, b/2)
Since the chord is bisected by the x-axis, the y-coordinate of the midpoint is zero.
Therefore, b/2 = 0 ⇒ b = 0
Now, we have two cases to consider: ab = 0, which means either 'a' or 'b' is zero.
Case 1: b = 0
Since b = 0, we have the point (a, 0) lying on the circle equation.
Substitute b = 0 in the circle equation:
x² + y² = ax + 0y
x² + y² = ax
Now, substitute this into the circle equation:
x² + y² = x(a)
Rearrange it:
x² - ax + y² = 0
We know that this equation represents a circle centered at (a/2, 0) with radius (a/2).
Applying the formula for radius of a circle, we have:
Radius = √((a/2)² + 0²)
Radius = a/2
Since the circle has a radius of a/2 and is centered at (a/2, 0), this implies that the x-coordinate of the center is the radius.
So we have:
a/2 = a ⇒ a = 2a/2 ⇒ 2a - 4 = 0 ⇒ a = 4/2 = 2
Therefore, in this case, a = 2 and b = 0.
Substituting these values into a² = 8b²:
(2)² = 8(0)²
4 = 0
Since 4 does not equal 0, this case does not satisfy the equation a² = 8b².
Case 2: a = 0
Since a = 0, we have the point (0, b) lying on the circle equation.
Substitute a = 0 in the circle equation:
x² + y² = 0x + by
x² + y² = by
Now, substitute this into the circle equation:
x² + y² = y(b)
Rearrange it:
y² - by + x² = 0
We know that this equation represents a circle centered at (0, b/2) with radius b/2.
Applying the formula for the radius of the circle, we have:
Radius = √(0² + (b/2)²)
Radius = b/2
Since the circle has a radius of b/2 and is centered at (0, b/2), this implies that the y-coordinate of the center is the radius.
So we have:
b/2 = b ⇒ b = 2b/2 ⇒ 2b - 4b = 0 ⇒ -2b = 0 ⇒ b = 0
Therefore, in this case, a = 0 and b = 0.
Substituting these values into a² = 8b²:
(0)² = 8(0)²
0 = 0
Since 0 equals 0, this case satisfies the equation a² = 8b².
To sum up, when ab = 0, if the chords bisecting the circle are also bisected by the x-axis, then a² = 8b² holds true when a = 0 and b = 0.
Hence, we have successfully proven that a² = 8b².
To prove that a^2 = 8b^2, you need to follow a logical series of steps. Let's break it down.
1. Start with the equation of the circle: x^2 + y^2 = ax + by. Rearrange it a bit to get y^2 - by = ax - x^2.
2. Consider two distinct chords drawn from the point (a, b) on the circle. Let's call the points where these chords intersect the circle as P, Q, R, and S. So, we have PQ and RS as the two chords.
3. Since the chords are bisected by the x-axis, the y-coordinates of both midpoints are zero. Let's call the midpoints of PQ and RS as M and N, respectively. Therefore, we have M( (x_P + x_Q)/2, 0) and N( (x_R + x_S)/2, 0).
4. To find the coordinates of M and N, we need the coordinates of P, Q, R, and S.
- P: Substitute y = ax - x^2 into the equation of the circle to get x^2 + (ax - x^2)^2 - b(ax - x^2) = 0. Solve for x, and you'll find two values for x.
- Q: Use the same steps as for finding P to get the second value of x.
- R: Substitute y = ax - x^2 into the equation of the circle to get x^2 + (ax - x^2)^2 - b(ax - x^2) = 0. Solve for x, and you'll find two new values for x.
- S: Use the same steps as for finding R to get the second value of x.
5. Once you have the x-coordinates of P, Q, R, and S, calculate the average of x-coordinates for PQ and RS.
- Average of x-coordinates of PQ: (x_P + x_Q)/2 = (x_1 + x_2)/2
- Average of x-coordinates of RS: (x_R + x_S)/2 = (x_3 + x_4)/2
6. Now, since M and N are on the x-axis, both x_1 + x_2 and x_3 + x_4 must be equal to zero.
7. Solve the equations x_1 + x_2 = 0 and x_3 + x_4 = 0 to find the relationships between the values of x_1, x_2, x_3, and x_4.
8. Once you have x_1, x_2, x_3, and x_4, substitute them into the equation of the circle (y^2 - by = ax - x^2) to get the corresponding y-coordinates.
9. Use the midpoint formula to calculate the coordinates of M and N.
10. Equate the x-coordinates of M and N to find a relationship between x_1, x_2, x_3, and x_4.
11. Simplify the relationship to obtain an equation in terms of a and b.
12. Compare the equation obtained in step 11 to a^2 = 8b^2. If they are equal, then you have proved that a^2 = 8b^2.