Question

Find the other end of the diameter of the circle x square + y square -8x -8y +27 =0
, if one end of it is
(2, 3)

Answer 1

Given :

Equation of circle

S = x² + y² - 8x - 8y +27

(2, 3) is one end of diameter

To Find :

Other end of the diameter

Solution :

•let coordinates of other end of circles are ( h , k )

•Given equation of circle is

x² + y² - 8x - 8y +27 = 0

x² - 8x + y² - 8y = -27

Adding 32 on both sides

(x² - 8x + 16) + (y² - 8y + 16 ) = -27+32

( x - 4 )² + ( y - 4 )² = 5

•Hence , centre of Circle lies on point ( 4 , 4 ) & radius of circe is √5 units

•Since centre divides diameter in two equal parts ,

By section Formula

x = ( m2x1 + m2x1 )/ ( m1 + m2 )

y = ( m2y1 + m2y1 )/ ( m1 + m2 )

•Where (x,y) are coordinates of centre & (x1,y1) , (x2,y2) are coordinates of diameter

So ,

4 =[ (1)(h) + (1)(2) ]/(1+1)

4 = ( h + 2 )/2

8 = h + 2

h = 6

Similarly ,

4 =[ (1)(k) + (1)(3) ]/(1+1)

4 = ( k + 3 )/2

8 = k + 3

k = 5

•Hence , Coordinates of center of circle are ( 6 , 5 )

Answer 2

Answer:

xsquare+ysquare+8x-5y+25