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If a circle passes thr . (2,2) (9,9) and touches x axis at p . Find op (where 0 is origin) =

If a  circle passes thr . (2,2) (9,9) and touches x axis at p . Find op (where 0 is origin) =
 

Grade:11

1 Answers

Rajat
213 Points
4 years ago
Let the centre of the circle be ( a, b).
Distance of the centre of the circle from the x-axis is equal to b units.
the circle touches the x-axis thus the radius of the circle is equal to the distance of the centre of the circle from x axis= b units.
So, eqn of the circle is 
(x-a)2+(y-b)2= b2
 
The circle passes through (2,2) and (9,9), thus 
(2-a)2+(2-b)2=b2
or, 4+a^2-4a+4+b^2-4b=b^2
Or, 8+a^2-4a-4b=0 
or, b=(a^2-4a+8)/4
 
(9-a)^2+(9-b)^2=b^2
Or, 162+a^2-18a-18b=0
Or, b= (162+a^2-18a)/18
 
So, 9(8+a^2-4a)=2(162+a^2-18a)
Or, 7a^2=252
Or, a^2=36 
Therefore a=±6
Co-ordinate of point P is equal to (±6,0) so, OP =6 units
 
 
 

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