How do you find the point or points of intersection of a straight line and a circle?
Given the equation of the straight line is: x+y=9 and the equation of the circle is: (x-2)2+(y-3)2=16 , find the point or points of intersection? The equation x+y=9 can be rearranged to make x=9-y. This equation equal to x can then be substituted into the equation of the circle to form (9-y-2)2+(y-3)2=16. If you simplify and square the bracket you get 2y2-20y+58=16. This can be simplified to 2(y-3)(y-7)=0. Thus y-7=0 so y=7 and y-3=0 so y=3. These are the y axis intersections of the circle and the line. At this point they both have the same x value and y value and therefore by definition touch. As there is two points the circle and line make contact it is an intersection, instead of a tangent (one point of contact). The x value can be found by substituting both y values into the equation of the line. For y=7 it yields the x value of 2 and for y=3 it yields the x value of 6. The x and y values of the coordinate can then be substituted into the eqaution of the circle to confirm that no algebraic error has been made.
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