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Find the equation of a circle tangent to the line 3x - 4y = 34 at the point (10, -1) and also
tangent to the line 4x + 3y = 12 at the point (3, 0).
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This problem is NOT an ORDINARY problem : it is a SPECIAL.
It is a special, since it is OVER-DEFINED.
It is clear that the center of the circle lies at the intersection of perpendicular lines to the given lines
at the given points;
but when we will find this intersection, we should CHECK that the distance is the same from the intersection
point to both given points.
So I came to assist you in making all necessary steps.
(1) The line perpendicular to 3x - 4y = 34 is 4x + 3y = const (you swap x and y and change the sign in the left side at one term).
In order for the point (10,-1) would belong to this perpendicular line, the const must be const = 4*10 + 3*(-1) = 40 - 3 = 37.
So, first perpendicular line is 4x + 3y = 37.
(2) The line perpendicular to 4x + 3y = 12 is 3x - 4y = const (you swap x and y and change the sign in the left side at one term).
In order for the point (3,0) would belong to this perpendicular line, the const must be const = 3*3 - 4*0 = 9 - 0 = 9.
So, second perpendicular line is 3x - 4y = 9.
(3) To find the intersection point, you should solve this system of two equations
4x + 3y = 37 (1)
3x - 4y = 9 (2)
The intersection point is (by the determinant method)
x = = 7; y = = 3.
(4) So, the intersection point and the center of the circle is the point (x,y) = (7,3).
The distance to the point (10,-1) is = = = 5.
The distance to the point (3,0) is = = = 5.
(5) The distances are the same, so the problem is posed correctly.
(6) The standard form equation of the circle is + = .
Solved.