Tiêu đề MSTE 12 Solutions.pdf ✅

Subtract the first and second equations −6x − 2y − 2 = 0 3x + y = −1 Subtract the second and third equations 8x + 8y − 8 = 0 x + y = 1 From the two equations, x = −1, y = 2 (−1, 2) ▣ 4. Find the equation of the circle having its center on the line 4x − y = 7 and passing through the points (−2,4) and (5,5). [SOLUTION] Let (h,k) be the center of the circle. 4h − k = 7 Since the circle passes through (-2,4) and (5,5), then the equation of the perpendicular bisector of the segment whose endpoints are the given points passes through the center of the circle. Midpoint of the segment −2 + 5 2 = 3 2 4 + 5 2 = 9 2 Slope of the segment m = 5 − 4 5 − (−2) = 1 7 Slope of the perpendicular bisector m⊥ = −7 Equation of perpendicular bisector y − 9 2 = −7 (x − 3 2 ) 2y − 9 = −14x + 21 14x + 2y = 30 7x + y = 15 Since it passes through the center, 7h + k = 15

20x − 9y = 34 The slope of the line is m = 20 9 therefore the slope of the perpendicular line is m⊥ = − 9 20 The equation of the line, using point-slope form is y − (−3) = − 9 20 (x − 2) 20y + 60 = −9x + 18 9x + 20y + 42 = 0 ▣ 7. Find the equation/s of the circle/s tangent to both axes and containing the point (−8, −1). [SOLUTION] Since the point is within the third quadrant, then the circle must also be located at the third quadrant. If the radius of the circle is r and the circle is tangent on both axes, then the coordinates of the center, at the third quadrant, is (−r, −r). The equation of the circle is (x + r) 2 + (y + r) 2 = r 2 Since it passes through (−8, −1), (−8 + r) 2 + (−1 + r) 2 = r 2 (r 2 − 16r + 64) + (r 2 − 2r + 1) = r 2 r 2 − 18r + 65 = 0 (r − 5)(r − 13) = 0 r = 5, 13 Therefore, there are 2 possible answers. (x + 5) 2 + (y + 5) 2 = 25 (x + 13) 2 + (y + 13) 2 = 169