Are you sure that you can solve this !!!!! (Problem related to circles) Part2
What are the cordinates of the center of a circle that is inscribed in a square made by the lines x 2 - 8 x + 12= 0 and y 2 - 14 y + 45= 0
Note: The coordinates are of the form ( m , n ) . Enter the answer as m + n .
The answer is 11.
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2 solutions
Nice solution @Arya
did the same way . is there any other way ?
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You could use derivatives to find the vertex (center of the parabolic function), which makes life easier. Makes it 2x-8=0, 2y-14=0.
By adding the two equations together, one obtains:
( x 2 − 8 x + 1 2 ) + ( y 2 − 1 4 y + 4 5 ) = 0 ;
or ( x 2 − 8 x + 1 6 − 4 ) + ( y 2 − 1 4 y + 4 9 − 4 ) = 0 ;
or ( x − 4 ) 2 + ( y − 7 ) 2 − 8 = 0 ;
or ( x − 4 ) 2 + ( y − 7 ) 2 = 8 .
Thus, the circle's center is ( m , n ) = ( 4 , 7 ) .
To solve this problem, you need to know some basic concepts of quadratic and coordinate geometry. from x 2 − 8 x + 1 2 we get ( x − 2 ) ( x − 6 ) from y 2 − 1 4 y + 4 5 we get ( y − 9 ) ( y − 5 ) So, the square is formed from the interection of these lines. The following points of sq. ABCD are ( 2 , 9 ) ( 6 , 9 ) ( 6 , 5 ) ( 2 , 5 ) We find the side of the sq. Which is 4 , it is also equal to the diameter of the incentre. since the circle touches the midpoints of the sides of sq. (Basic observation) From the midpoint's coordinates and the radius 2 , we can find out the centre's coordinates which comes to be ( 4 , 7 )