What is the point of intersection and distance of line 3x+4y=10 and (-1,-3)?

Question

Find the equation of the circle with its center at (-1,-3) and tangent to the line through the point (-2,4) and (2,1). 

Answer 1

write the line 3x + 4y = 10 in slope-intercept form:  y = -(3/4)x + (5/2) = mx + b, so that m = -(3/4);  next, consider the line, y = Mx + B, passing through (-1,-3), perpendicular to the line 3x + 4y = 10, intersecting that line at (c,d);  since the line y = mx + b is perpendicular to the line y = Mx + B, we have that M = -(1/m) = 4/3; next, calculate

M = 4/3 = (d - (-3))/(c - (-1)) = (d + 3)/(c + 1);  in the preceding equation, cross multiply, to obtain: 

4(c + 1) = 3(d + 3), or 4c + 4 = 3d + 9, or 4c - 3d = 5;  next, use the fact that (c,d) lies on the line 3x + 4y = 10, meaning that 3c + 4d = 10;  now, we two equations in two unknowns, namely:

 

            4c - 3d = 5

            3c + 4d = 10

 

multiply the top equation by (-3), and the bottom equation by 4, yielding the following:

 

            -12c + 9d = -15

              12c + 16d = 40

 

adding the two equations above yields:  25d = 25, which yields d = 1; 

using this value of d in 4c - 3d = 5 yields the following:  4c - 3 = 5, which yields 4c = 8, which yields c  = 2; so that the required point of intersection is (c,d) = (2,1)

 

next, used the distance formula to find the (perpendicular) distance from the point (-1,-3) to the line 3x + 4y = 10:

 

D^2 = [(2 - (-1)]^2 + [(1 - (-3)]^2 = (3^2 + 4^2) = 9 + 16 = 25, which implies that D = 5, where D represents the perpendicular distance of the point (-1,-3) to the line 3x + 4y = 10

 

note that the question that asks you to find the equation of the circle with its center at (-1,-3) and tangent to the line passing through  the points (-2,4) and (2,1), is precisely the question just answered above, because of the following:

the slope of the line passing through the points (-2,4) and (2,1) is (4 - 1)/(-2 - 1) = 3/(-4) = -(3/4) = m;  also, using the Point-Slope form of the equation of the straight line passing through the pointe (-2,4) and (2,1) yields the following:

 

      y - 1  = -(3/4)(x - 2), or (multiplying through by) 4y - 4 = -3x + 6, or 3x + 4y = 10,

 

which is precisely the equation of the line originally given;  so, recalling that the radius of the circle in question is perpendicular to the line tangent to the circle at the point (c,d) = (2,1), we have that the equation of the circle in questions is given by the following:

 

      (x - (-1))^2 + (y - (-3))^2 = D^2; or (x + 1)^2 + (y + 3)^2 = 5^2; or (x + 1)^2 + (y + 3)^2 = 25

Answer 2

Any radius drawn to a tangent line forms a right angle.  This means that

(slope of radius) = -1/(slope of tangent).

 

We can use this to develop a strategy:

(1)  Find the equation of the tangent line.

 

(2)  Find the equation of the line containing the radius, using the negative reciprocal slope of the tangent line, and the center of the circle as a point on the line.

 

(3)  Find the point of intersection of these lines (call it point P).  This is where the tangent line and radius touch.

 

(4)  Use the distance formula to find the distance from the center of the circle to point P.  This gives us the length of the radius.

 

(5)  substitute known values for the circle center (h,k) and radius (r) into the equation for a circle:

        (x-h)² + (y-k)² = r²

 

Follow these steps and you'll have the answer.