What are the value S of Y if the points A (- 1 Y and B 57 lie on a circle with Centre 0 2 3y )?

Text Solution

Solution : Here, `" "OA=OB" "`(radii of a circle) <br> `rArr" "OA^(2)=OB^(2)` <br> `rArr" "(2+1)^(2)+(-3y-y)^(2)=(2-5)^(2)+(-3y-7)^(2)` <br> `rArr" "9+16y^(2)=9+9y^(2)+42y+49` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NTN_MATH_X_C07_S01_056_S01.png" width="80%"> <br> `rArr" "7y^(2)-42y-49=0` <br> `rArr " "y^(2)-6y-7=0` <br> `rArr" "y^(2)-7y+y-7=0` <br> `rArr" "y(y-7)+1(y-7)=0` <br> `rArr" "y-7=0" "or" "y+1=0` <br> `rArr" "y=7 " "or " "y=-1` <br> Now, the co-ordinates of centre O are either (2, -21) or (2, 3) when centre is O(2, -21), then <br> `" ""radius"=OB` <br> `" "=sqrt((2-5)^(2)+(-21-7)^(2))` <br> `" "=sqrt(9+784)=sqrt(793)` units <br> When centre is O(2, 3), then <br> `" ""radius"=OB` <br> `" "=sqrt((2-5)^(2)+(3-7)^(2))` <br> `" "=sqrt(9+16)=sqrt(25)=5` units

A and B are the two points that lie on the circle and O is the centre of the circle.Therefore, OA and OB are the radii of the circle.

Using the distance formula, we have:

`OA=sqrt((-1-2)^2+(y+3y)^2)=sqrt(9+16y^2)`

`OB=sqrt((5-2)^2+(7+3y)^2)=sqrt(9+(7+3y)^2)`

Now, OB = OA              (Radii of the same circle)

`sqrt(9+(7+3y)^2)=sqrt(9+16y^2)`

9+(7+3y)2=9+16y^2 (squaring both the sides)

49+9y2+42y=16y2

⇒7y2−42y−49=0

⇒y2−6y−7 =0

⇒y2−7y+y−7=0

⇒(y−7)(y+1)=0

⇒y−7=0  or y+1=0

⇒y=7 or y=−1

When y = 7:
Radius of the circle, `OA=sqrt(9+16y^2)=sqrt(9+16×49)=sqrt(793)`

When y = −1:
Radius of the circle, `OA=sqrt(9+16y^2) =sqrt(9+16×1)=sqrt(25)=5`