Question

If the points(a^2,2a),(1,-2)and(2,-1)never lie on a unique circle, then find the value of a

Answer 1

a = 3 then points (a²,2a) ,(1,-2)and(2,-1) never lies on unique circle

Step-by-step explanation:

Complete Question :

If a>0 and (a²,2a) ,(1,-2) and (2,-1) never lie on a circle, then possible value of a is

a) 1

b) 2

c) 3

d) 4

if a²,2a lies on line passing through (1,-2) and (2,-1)    & ≠ (1,-2) or (2,-1)

Then it will never lie on that circle which passes through (1,-2) and (2,-1)

slope of line passing through (1,-2) and (2,-1)  =  ( -1 - (-2))/(2 - 1)

= 1 /1

= 1

y = x + c

-1 = 2 + c

=> c = -3

y = x - 3

if a² & 2a lies on this

2a = a² - 3

=> a² - 2a - 3 = 0

=> a² + a - 3a - 3 =0

=> a(a + 1) - 3(a + 1) = 0

=> a = 3 or a = -1

a > 0

=> a = 3

the value of a = 3 then points (a²,2a) ,(1,-2)and(2,-1) never lies on unique circle

How many circles can be drawn through three collinear points?

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Answer 2

The value of 'a' is -1 or 3.

Step-by-step explanation:

Consider the provided points.

We need to find the value of a so that the three points never lie on a unique circle.

3 points never lie on a unique circle if they are collinear.

So first find the equation of line passing through (1,-2) and (2,-1).

Find the equation of line by using the formula: $(y-y_1)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$(y-(-2))=\frac{-1-(-2)}{2-1}(x-1)$

$y+2=1(x-1)$

$y=x-1-2$

$y=x-3$

Now substitute $x=a^2\ and\ y=2a$

$2a=a^2-3$

$a^2-2a-3=0$

$a^2+a-3a-3=0$

$a(a+1)-3(a+1)=0$

$(a+1)(a-3)=0$

Hence, the value of a=-1 or 3.

#Learn more

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