The diameter AB. A(6;4) and B(0;-4) is given. CBA equals delta

Question

Please give full calculations and reasonings. Please help me urgently 

Answer 1

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4.1) The equation of a circle is of the form (x-h)²+(y-k)²=r² where (h,k) is the centre and r is the radius. The centre is the midpoint of the diameter AB=((6+0)/2,(4-4)/2)=(3,0). h=3 and k=0.

AB=2r, so 2r=√((6-0)²+(4-(-4))²)=√(36+64)=10, r=5

Equation of the circle is (x-3)²+y²=25

4.2) When y=0 (x-axis) (x-3)²=5², x-3=±5, so x=3±5=8 or -2. C is in the negative region of x so C is (-2,0).

4.3) Slope (gradient) of AB=(4-(-4))/(6-0)=8/6=4/3. The slope of the perpendicular CD is the negative reciprocal of this = -¾. CD passes through C(-2,0) so the equation of CD is y-0=-¾(x-(-2)), y=-3x/4-3/2. This can be written: 3x+4y+6=0.

4.4) The slope of AB is 4/3 and passes through B(0,-4) so the equation of AB is y=4x/3-4. M is the intersection of AB and CD, so we equate the y's: -3x/4-3/2=4x/3-4.

Multiply through by 12: -9x-18=16x-48, 25x=30, x=30/25=6/5. y=4(6/5)/3-4=8/5-4=-12/5. So M is (6/5,-12/5).

4.5) D is the intersection of CD and the circle, so we use the equations of each to find the coordinates of D. From 4.3 we can substitute for y in the equation of the circle (4.1):

(x-3)²+(3x/4+3/2)²=25, x²-6x+9+9x²/16+9x/4+9/4=25,

16x²-96x+144+9x²+36x+36=400 (multiplying through by 16),

25x²-60x-220=0=5x²-12x-44=(x+2)(5x-22) (we already knew one factor must x+2 because of C(-2,0)). So x=22/5, y=-¾x-3/2=-¾(22/5)-3/2=-33/10-3/2=-24/5.

D is (22/5,-24/5) or (4.4,-4.8)

Mark the centre of the circle O(3,0) and join C to O, CO=BO=AO=5 (radius). Length CM=√((6/5-(-2))²+(12/5)²)=√((16/5)²+(12/5)²)=√(256+144)/5=4. So MO=√(CO²-CM²)=√(25-16)=3.

Therefore, AM=AO+MO=5+3=8. tanÂ=CM/AM=4/8=½, and:

4.6.1) Â=arctan(½)=26.6° approx.

4.6.2) D̂=Â=26.6° because these angles are in the same segment standing on BC.

4.6.3) ∠ACB=90° (angle in semicircle is a right angle), so Â+θ=90°, θ=90-26.6=63.4°.

4.7.1) Since A, B, and C are already on the circumference of the circle, E must also be on the same circumference. Since it's on the y-axis, its x coordinate must be zero, so put x=0 in the equation of the circle: 9+y²=25, y²=16, y=4 (we can't use -4 because that's B). E is (0,4).

4.7.2) Triangle ACO is isosceles because two arms AO and CO are radii. ∠ACO=Â=26.6°. ∠AOC=180-2Â. Reflex angle AOC=360-(180-2Â)=180+2Â. E is the angle at the circumference and AOC is the angle at the centre, so ∠AEC=90+Â=90+26.6=116.6°.

Please note that there may be other ways and reasons to derive these answers.