the centre of the circle circumscribing the square whose 3 sides are 3x+y=22 , x-3y=14 and 3x+y=62 is
Answers
Answer:
Given 3 lines. Two simultaneous equations yields two points of intersection A(8,-2) and B(20,2). Let the corners be A,B,C and D. D and B are equidistant from A. Using Pythagoras the squared distance from A to B is 160. The length squared between A and D is 160.
(b+2)^2 +(a-8)^2 = 160. We know (a,b) satisfies eq. y = -3x+22. So b = -3a+22
Replace b with -3a+22 in above equation.
(-3a+24)^2 + (a - 8)^2 = 160
a^2 -16a +48=0 giving a = 12 or a = 4. a = 4 works. b = -3(4)+22 = 10.
Now get mid point of diagonal D(4,10) to B (20,2) = (12,6) the Centre of the circumcircle.
Answer (12,6
Answer:
Lets name the three lines and state their equations:
Line L1:3x+y=22
Line L2:x−3y=14
Line L3:3x+y=62
The equation of the line L4 is not specified. So there could be two different squares possible, because L4 could be on either side of L2.
From the equations, we see that L1 is parallel to L3 . We can say that because both have the expression 3x+y on the left hand side. The constant on the right hand side is different. That means that the y-intercept of the two parallel lines are 22 and 62 . The y-intercept of the mid-line between L1 and L3 has to be 42 . So the equation of this mid-line has to be
3x+y=42
The intersection point A between L1 and L2 can be found to be (8,−2) by solving the pair of equations of L1 and L2 simultaneously. Similarly, the intersection point B between L2 and L3 is found to be (20,2) . AB is one side of the square and we find that mid-point of AB is (14,0) . The length of AB is 410−−√ , which means that the half length of the square is 210−−√ .
The centre of the circumscribing circle has to lie on 3x+y=42 and has to be at a distance 210−−√ from (14,0) . This is enough information for us to be able to find the coordinates of the two possible centres of the circumcircles.
Since the circumcentre has to lie on 3x+y=42 , the coordinates of the circumcentre can be expressed as (x,−3x+42) . Since its distance from (14,0) is 210−−√ , we have
(x−14)2+(−3x+42)2=(210−−√)2
solving which we find the coordinates of the two centres to be (12,6) and (12,−6)
Hope it helps