Question

The equation of the circle which touches the line y=5 and that passes through (-1,2) and (1,2) is __________

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Answer 1

Given : circle  touches the line y=5 and passes through (-1,2) and (1,2)

To Find : Equation of circle

Solution

circle passes through (-1,2) and (1,2)

center of circle will  be at perpendicular bisector of (-1,2) and (1,2)

Slope of  (-1,2) and (1,2)  = 0

Mid point of  (-1,2) and (1,2)   = ( 0 ,2 )

Hence center of circle will lie on y axis

Let say ( 0 , y)

circle  touches the line y=5

Then   then radius  = |  y - 5 |

(0 - 1)² + ( y - 2)²  = | y - 5 |²

=> 1  + y²  - 4y + 4 = y²  -10y + 25

=> 6y = 20

=> y = 10/3

Center of circle = ( 0 , 10/3)

Radius = | y - 5 |  = | 10/3 - 5|  = 5/3

x²  + ( y - 10/3)²  = (5/3)²

=> 9x²   + (3y - 10)²  = 25

=> 9x² + 9y²  - 60y   + 100 = 25

=> 9x² + 9y²  - 60y  + 75 = 0

=>  3x² + 3y²  - 20y  + 25 = 0

The equation of the circle which touches the line y=5 and that passes through (-1,2) and (1,2) is 3x² + 3y²  - 20y  + 25 = 0  

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Answer 2

$9x^{2} +9y^{2} - 60y + 75 = 0$

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