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Suppose a circle passes through (2,2) and (9,9) and touches the X - axis at P . If O is the origin then OP=

Suppose a circle passes through (2,2) and (9,9) and touches the X - axis at P . If O is the origin then OP=

Grade:12th pass

1 Answers

Arun
25750 Points
6 years ago
Write the general equation of a circle with center (h,k) and radius r: 
(x - h)² + (y - k)² = r² 

We know 3 points on this circle, the two points given and the tangent point on the x-axis: 
(2,2) 
(9,9) 
(h,0) 

Plug in each of these points: 
(2 - h)² + (2 - k)² = r² 
(9 - h)² + (9 - k)² = r² 
(h - h)² + (0 - k)² = r² 

We get k² = r² from the last equation, so let's replace that in the first two equations: 
(2 - h)² + (2 - k)² = k² 
(9 - h)² + (9 - k)² = k² 

Expand both out: 
4 - 4h + h² + 4 - 4k + k² = k² 
81 - 18h + h² + 81 - 18k + k² = k² 

Cancel k²: 
4 - 4h + h² + 4 - 4k = 0 
81 - 18h + h² + 81 - 18k = 0 

Simplify: 
h² - 4h + 8 - 4k = 0 
h² - 18h + 162 - 18k = 0 

We want to get rid of k, so multiply the first equation by 9 and the second equation by 2: 
9h² - 36h + 72 - 36k = 0 
2h² - 36h + 324 - 36k = 0 

Subtract the second equation from the first: 
7h² - 252 = 0 

Divide both sides by 7: 
h² - 36 = 0 
h² = 36 
h = ±√36 

But we only care about the positive solution since the negative wouldn't make sense. 

Answer: 
h = 6 

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