1)The center of a circle is (2a,2a-7) find the value of 'a' of the circle passes through the point (11,-9) and has diameter 10√2. 2)Two vertices of equilateral triangle are (3,0) and (6,0) find the third vertex 3)Find the ratio in which the linesegment joining A(1/2,3/2) B(1,-5). 4)If point A(0,2) is equidistant from the point B(3,P) C(P,5) find 'P' also find the length of AB. [Don't spam]
Answers
The centre of the circle (0)=(2a,a−7)
Point (P)=(11,−9)
diameter (D)=102 units
Also given, to find out the values of a
from figure, we get OP=7 units
∴ OP=(x2−x1)2+(y2−y1)2²
O=(2a,a−7), P(11,−9)
⇒ 7=(11−2)2+(−9−(a−7))2²
7=121+4a²−44a+(−9−a+7)²
7=121+4a²−44a+a2+4a+4
7=5a²−40a+125
so on both sides, we get,
49=5a²−40a+125
⇒5a²−40a+76=0
We know that 2a−b±b2−4ac is used to findroots for quadratic equation
we get
5a²−40a+76=0;2a−b±b2−4ac
b=−40;a=5;c=76
⇒2×5−(−40)±(−40)2−4(5)(76)=1040±1600−1520
=1040±80=1040±45=104(10±5)=52(10±5)
∴a==52(10+5);a=52(10−5)
a=10.89=11;a=3.10=3
∴a=11,3
Hope this helps you
Please mark as brainliest
1. Given,
centre of circle = (2a, a-7)
Circle passes through point (11, -9)
Diameter of circle = 10√2 units
According to the prblm,
10√2/2 = √[(11 - 2a)² + (-9 - a + 7)²]
√50 = √[121 + 4a² - 44a + 4 + a² + 8a]
50 = 125 + 8a² - 32a
5a² - 40a + 75 = 0
5a² - 15a - 25a + 75 = 0
(5a - 25)(a - 3)
a = 3 or 5
2. Let, the third vertex be (x, y)
As the triangle is equilateral,
all sides of it are equal.
According to the prblm,
√[(x - 3)² + (y - 0)²] = √[(x - 6)² + (y - 0)²]
x² + 9 - 6x + y² = x² + 36 - 12x + y²
6x = 27
x = 27/6 = 9/2
√[(x - 3)² + (y - 0)²] = √[(6 - 3)² + (0 - 0)²]
x² + 9 - 6x + y² = 9
(9/2)² - 6(9/2) = -y²
81/4 - 27 = -y²
-27/4 = y²
y = 3√3/2
Therefore, third vertex = (9/2, 3√3/2)
3. Refer the above attachment.
4. Given, AB=AC
= √[(0-3)² + (2-p)²]
= √[(0-p)² + (2-5)²]
= 9 + (2-p)² = p² + 9
= 4 + p² - 4p = p²
4p =4
p = 1.
Substitute of p-value in AB
Length of AB = √[(0-3)² + (2-1)²]
= √(9 + 1)
= √10
