Circle Intercept
A circle with y -intercept 1 1 passes through the point ( 9 , 8 ) , and the equation of the circle is its own inverse.
What is the other y -intercept of the circle?
The answer is -7.
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2 solutions
How did you get y=3x-4
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A perpendicular bisector between two points is a line that passes through its midpoint with an opposite reciprocal slope.
The slope between ( 0 , 1 1 ) and ( 9 , 8 ) is m = x 2 − x 1 y 2 − y 1 = 9 − 0 8 − 1 1 = 9 − 3 = − 3 1 , so its opposite reciprocal is m = 3 .
The midpoint between ( 0 , 1 1 ) and ( 9 , 8 ) is M = ( 2 x 1 + x 2 , 2 y 1 + y 2 ) = ( 2 0 + 9 , 2 1 1 + 8 ) = ( 2 9 , 2 1 9 ) .
The equation of a line that has a slope of m = 3 and passes through the midpoint ( 2 9 , 2 1 9 ) is y − 2 1 9 = 3 ( x − 2 9 ) which simplifies to y = 3 x − 4 .
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Hey sorry for a doubt again, but at the last step how did you get x-(9/2) I understand equation of a line is y=mx+c, where m and c are slope and y intercept respectively.
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@Aman Thegreat – I used the point-slope form of the line, which is y − y 1 = m ( x − x 1 ) , where m is the slope, and ( x 1 , y 1 ) is the point it passes through. But you can also arrive at the same equation using the slope-intercept form of the line y = m x + b : substitutiting m = 3 , x = 2 9 , and y = 2 1 9 gives 2 1 9 = 3 ( 2 9 ) + b , and solving for b gives b = − 4 , and replacing m and b back in gives y = 3 x − 4 .
Since the circle is its own inverse, its center must lie on the line y = x . Thus, the equation corresponding to the graph of the circle can be written as ( x − a ) 2 + ( y − a ) 2 = r 2 , for some real numbers a , r . Since both ( 0 , 1 1 ) and ( 9 , 8 ) are on the circle, we have
a 2 + ( 1 1 − a ) 2 2 a 2 − 2 2 a + 1 2 1 1 2 a a = ( 9 − a ) 2 + ( 8 − a ) 2 = 2 a 2 − 3 4 a + 1 4 5 = 2 4 = 2 .
So, the equation of the circle becomes ( x − 2 ) 2 + ( y − 2 ) 2 = r 2 . One of the y -intercepts of this circle is given to be 11, so the other y -intercept will be at the solution to ( y − 2 ) 2 = ( 1 1 − 2 ) 2 , where y = 1 1 . This can easily be seen to be y = − 7 , and we are done.
Great approach and solution!
The locus of points of the centers of all possible circles that pass through ( 0 , 1 1 ) and ( 9 , 8 ) is the perpendicular bisector of these points, which is y = 3 x − 4 .
For an equation to be its own inverse, it must be symmetric in y = x . A circle is symmetric in any line through its center, so the locus of points of the centers of all possible circles that are its own inverse is y = x .
The center of our circle must then be the solution to these two equations which is ( 2 , 2 ) .
One y -intercept is 1 1 , which is 9 units above its center, and since a circle is symmetric around its center, the other y -intercept must be 9 units below it, which is − 7 .