How to find the intersection of a straight line and a circle algebraically?

Question

Find the set of values of m for which the line y = mx + 1 intersects the circle (x - 7)² + (y - 5)² = 20 at two

distinct points.

I understand the question and can find correct values of m for which they are true however can only come to a correct set of values for m by trial and error. I feel like this is a long and tedious way of going about it and wouldn’t have time to do it like this in an exam situation….which makes me think I’m doing it the hard way or missing some vital shortcut? I can complete the same type of question with a real number instead of m however can’t seem to solve it algebraically. Any help would be very much appreciated :)

Answer 1

one easy value of m is 0, then the intersection points are (5,1) and (9,1)

scroll down to the bottom to see the calculations, another is m=1, with about (2.7, 3.7) and (8.3, 9.3)

(x-7)^2 + (y-5)^2 = 20

(x-7)^2 + (mx+1 -5)^2 = 20

solve for x in terms of m

or solve for m in terms of x

It helps to graph the equations, the circle has center (7,5) with radius 2sqr5 = about 4.4

the line has y intercept = 1 and slope = m = between the slopes of tangent lines to the circle, . Graph the circle and potential lines from (0,1) tangent to the circle. one tangent line has positive slope. The other is slightly negative

the slope = y' =(12-x)/y =m

(x^-7)^2 + (y-5)^2 = 20

x^2 -14x +49 + y^2 -10y +25 = 0

take the derivative with respect to x

2x -14 +2yy' -10y' =0

x-7 +yy' -5y' =0

(y-5)y' =7-x

y' =(7-x)/(y-5)

which = slope = m in the equation y=mx+1

find the points of tangency and find the slope of each line through them to the point (0,1) those two slopes are the upper and lower bounds for m. m can be anywhere in between, as those lines will all intersect the circle. Any slope larger or less than those upper & lower bounds will not intersect the circle

if you just want two values of m that intersect the circle, one is m=0

it intersects the circle at (5,1) and (9,1)

another is m=1 with intersection points about (2.7, 3.7) and (8.3, 9.3)

y=(0)x + 1 = 1

(x-7)^2 + (1-5)^2 = 20

x^2 -14x +49 +16 =20

x^2-14x +45 = 0

(x-9)(x-5) = 0

x = 5, 9

y =1,

the intersection points are (5,1) and (9,1)

for m=1

y=x+1

(x-7)^2 + (x+1-5)^2 = 20

expand and solve for x, there're 2 solutions

then add 1 to get y

x^2 -14x + 49 + x^2 -8x + 16 = 20

2x^2 -22x +45 = 0

use the quadratic formula

x =22/4 + or - (1/4)sqr(22^2 -4(2)(45)

x = about 8,28 or 2.72

y=x+1 = about 9.28 or 3.72

the intersection points are about (2.72, 3.72) and (8.28, 9.28)