Question

Plz guys hlp me wid this!!!

Answer 1

Hi Mate!!!

Let the co-ordinates of center of circle be ( h , k )

General equation of circle is

( x - h ) ² + ( y - k ) ² = R²

( 6 - h ) ² + ( -6 - k )² = R² ..... equation 1

( 3 -h )² + ( -7 - k )² = R² ......... equation 2

( 3 -h )² + ( 3 -k )² = R² ........ equation 3

Now solve these equations to find out (h,k)


Have a nice day...

Answer 2

Given points are A(6,-6), B(3,-7) and C(3,3)

Let the centre of the circle be D(x,y)

∴ DA = DB = DC

⇒ √(x - 6)² + (y + 6)² = √(x - 3)² + (y + 7)² = (x - 3)² + (y - 3)²

On squaring, we get

⇒ (x - 6)² + (y + 6)² = (x - 3)² + (y + 7)² = (x - 3)² + (y - 3)²

⇒ x² + 36 - 12x + y² + 36 + 12y = x² + 9 - 6x + y² + 49 + 14y = x² + 9 - 6x + y² + 9 - 6y

⇒ -12x + 12y + 72 = -6x + 14y + 58 = -6x - 6y + 18

(i)

-6x + 14y + 58 = -6x - 6y + 18

⇒ 20y = -40

⇒ y = -2.

(ii)

⇒ -12x + 12y + 72 = -6x + 14y + 58

⇒ -12x - 24 + 72 = -6x - 28 + 58

⇒ x = 3.

Therefore, centre of circle is D(3,-2).

Hope it helps!