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Problem Set 2, Problem # 2. Ellen Dickerson. Problem Set 2, Problem #2. Find the equations of the lines that pass through the point (1,3) and are tangent to the circle with equation x 2 + y 2 = 2 . A Circle with equation x 2 + y 2 = 2 has center point (0,0) and radius √(2)
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Problem Set 2, Problem # 2 Ellen Dickerson
Problem Set 2, Problem #2 Find the equations of the lines that pass through the point (1,3) and are tangent to the circle with equation x2 + y2 = 2
A Circle with equation x2 + y2 = 2 has center point (0,0) and radius √(2) draw/show point (1,3) and how there are two lines that go through (1,3) and are tangent to the circle.
Solving the Problem Algebraically • Midpoint (between the center of the original circle and the point (1,3)) • ((1+0)/2 , (3+0)/2) = ((1/2), (3/2)) • Distance (between the center of the original circle and the midpoint) • (√(12+32))/2 = (√10)/2 • Formula for the new circle • (x-(1/2)2 +(y-(3/2)2 = 5/2
Finding the points where the circles intersect (x-(1/2)2 +(y-(3/2)2 = 5/2 x2+y2 = 2 x2–x+y2-3y = 0 + -1(x2+y2)=-1 (2) -x-3y = -2 x = 2-3y Plug 2-3y into x in the equationx2+y2 = 2 (2-3y)2+y2 = 2 10y2-12y+2 = 0 (5y-1)(2y-2)= 0 5y-1= 0 and 2y-2= 0 5y=1 2y=2 y=(1/5) and y=1
Finding X (use formula x = 2-3y) y=(1/5) x=2-3(1/5) x=2-(3/5) x=(7/5) = 1.4 ( (7/5), (1/5 )) y=1 x=2-3(1) x=2-3 x=-1 (-1,1)
Finding the line ((7/5), (1/5 )) and (1,3) m=(1/5)-3/(7/5)-1 m=-(14/5)/(2/5) m=-7 y-3=-7(x-1) y-3 =-7x+(7) y= -7x +10 (-1,1), and (1,3) m=(1-3)/(-1-1) m=1 y-3=(1)(x-3) y-3=(x-1) y= x+2