Question

Find the equation of circle whose centre lies on the x-axis and passing through (-2,3) and (4,5)

Answer 1

Given:

The Circle is passing through (-2,3) and (4,5)

To find:

The equation of the circle.

Solution:

1) The center of the circle lies on the x-axis. So the coordinate of the center of the circle is (a,0)

2) The distance from center to points (−2,3) and (4,5) is equal , that would the radius of the circle.

So by the distance formula, we get

• (a+2)² +(0−3)²  =(a−4)²  +(0−5)²

• a²+4+4a+9 = a²+16-8a+25

• 12a=28

• a=7/3

3) The equation of the circle is:

• ​r²  =(x−h)²  +(y−k)²

where h,k are the coordinates of the circle.

• ​r²  =(4−7/3)²  +(5)²

• ​r²  =(5/3)²+5²

• r² = 25/9+25

• r²= 250/9

4)So the equation of the circle is given by:

• ​r²  =(x−h)²  +(y−k)²

• 250/10  =(x−7/3)²  +(y)²

• x²+49/9-14x/3+y² = 250/9

• x²-14x/3+y²= 250/9-49/9

• x²-14x/3+y² = 201/9

The equation of the circle is x²-14x/3+y² = 201/9.

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Answer 2

Answer:

centre lies on X- axis

A (-2,4) and B(4,5)

Required circle be X2 +Y2+2gx +2fy +c=o.........(1)

centre (-g,-f) lies on X- axis

f= 0

equation (1) passes through A(-2,3)

-4g+c+13=0........(2)

equation (1) passes through B(4,5)

8g+ +c+41=0.......(3)

from (2) and (3)

(3) -(2). 12g+28=0

g= -7/3

from (2)

28/3+C+13=0

C= -67/3

equation of the circle

X2+Y2-14/3x-67/3=0.