Flip a horizontally mirrored copy of this figure onto itself, to see that the diameter equals sqrt(8^2+1^1).

There is simpler way! Construct chord parallel to BD intersecting AC 2 below point A. This is symmetrical about O. AC is cut into 3 pieces: upper one = 2, mid one = 4, lower one = 2. Center 4 length is symmetrical about O, putting O 2 units above BD. Construct line through center parallel to AC to meet BD at P'. PP' = 1/2 by symmetry. Right triangle BP'O is formed, leg lengths 7/2, 2, hypotenuse = R. R² = (7/2)² + 2². R = √65/2.

We use an adapted orthonormal. P(0;0) A(0;6) B(-3;0) C(0;-2) D(4;0) The equation of the circle is x^2 + y^2 +a.x + b.y +c = 0 with a,b, c unknown. A is on the circle, so 9 -3.a +c = 0; B is on the circle, so 16 +4.a +c =0; C is on the circle, so 36 +6.b +c = 0; D is on the circle, so 4 -2.b +c = 0 The system is very easy to resolve and gives a = -1; b = -4; c = -12, so the equation of the circle is x^2 + y^2 -x -4.y -12 = 0 This equation is also (x -(1/2))^2 + (y -2)^2 = 12 +4 + (1/4) = 65/4. So the radius is sqrt(65)/0 and O(1/2; 2).

I seem to be something of an outlier with the way I do these: Either line may be shifyed, but I will opt to move the horizontal chord upwards, purely because the vertical is an even-numbered length and it makes the calculation slightly easier. Move the horizontal chord up to pass over the diameter.This gives chord sections of 4 and 4 for the vertical and (4+x) and (3+x) for the horizontal. x is the distance between the ends of the chord (after moving it) and the circumference. It is the same distance on both sides so can have the same variable name. 4*4=(x+4)(x+3) 16 = x^2 + 7x + 12 (we need to find the value of 2x and add it to the chord length 7. Quadratic formula: x^2 + 7x - 4 = 0 (-7+or-sqrt(49-4*1*-4))/2 = x (-7+or-sqrt(65))/2 = x. Actually, I can dispense with the /2, because I want 2x. 2x = sqrt(65) - 7, so the diameter is 7 + sqrt(65) - 7, which simplifies to sqrt(65). r = sqrt(65)/2 In decimal this is 4.031 (rounded).

why so difficult? as cords intersect exactly at the middle to the center point only one pythagoras is needed: 2**2 +3,5**2=r **2 direct result r=sqrt 65 divided by 2

@55dembel

22 күн бұрын

Exactly! Didn't even need to draw anything, just looked at the picture. for a minute)

Through O draw diameter perpendicular to AC. Then (r-1/2)(r+1/2)=(4)(4) so r^2-1/4=16. So r^2=65/4 giving r=sqroot65 /2

@larswilms8275

Ай бұрын

Nice solution. Next time though please explain how you got the 1/2 added and subtracted from the radius from. You now bring it as a big surprise. It comes from the distance from the O to the line AC. And can be calculated using line BD (BD = BP + DP = 3+4 = 7) The midpoint of BD is at the same distance from P as O from line AC, The midpoint is 7/2 from B, therefor the distance from the midpoint is 1/2

@johnbrennan3372

Ай бұрын

@@larswilms8275 Thanks for your reply. You worked out yourself very nicely how the 1/2 came about.That’s exactly how I worked it out myself.

three of the points form a triangle with apices on the circle - its easy to work out the circumcentre with no fuss...

The coordinates of the four points are obvious. A: ( -0.5 , 4 ) B: ( -3.5 , -2 ) C: ( -0.5 , -4 ) D: ( 3.5 , -2 ) r² = x² + y² Using Point A: r² = (-1/2)² + 4² r² = 1/4 + 16 r² = 65/4 r = (√65)/2

It seems there are lots of easier ways! Mainly just using the perp bisector of a chord passes through the centre, and Pythagoras of course.

@pk2712

13 күн бұрын

You are definitely correct ; but , doing it the easy way would only result in a video of about one minute duration .

Fun. Took the image, rotated 180deg and superimposed on itself. Produced a 1 x 4 rectangle centered symmetrically around circle center. r^2 = 16^2 + 0.5^2

R=√65/2

@Masterclass_Geometry

2 ай бұрын

great👌

Very nice problem! I think it can be solved in a somewhat simpler way though 🙂 Draw a line EF that is parallel to BD in the top half of the circle at distance 2 from A. By symmetry, the center of the rectangle EFDB is O. The vertical line through O intersects BD at point Q. The right-angled triangle BOQ has sides 3.5 (BQ) and 2 (OQ, because the distance from BD to EF is 4), so the radius of the circle, which is the length of BO, is the square root of 16.25.

@GeometryDose

3 ай бұрын

I agree there are indeed simpler ways to solve this for the radius. The main purpose I made this video was to show how to derive the formula (2R)^2= AP^2 + BP^2 + CP^2 + DP^2. But I can't put this whole formula on the title of the video because fewer people will click the video that way so I hope you understand.

@Emil_Avg

3 ай бұрын

If you want a fast solution you could for example calculate the area of triangle ABC and then use the formula which connects this area with the radius of the circumcircle.

It's much simpler than that. Draw OM ⟂ AC (M∈AC), so AM=CM=(6+2)/2 = 4. Then draw ON ⟂ BD (N ∈ BD), and AN = DN = (3+4)/2 = 7/2. Therefore PN= PD-DN = 4-7/2=1/2 and OM = PN = 1/2. From triangle OMA, R = OA = √(AM²+OM²) = √(4²+(1/2)² = √(4²+(1/2)²)=√(16+1/4) = √65 / 2. BTW, condition PC=2 is redundant. It isn't used here, but anyway it can evaluated, given that BP x PD = PC x PA.

answer is square root of 17. Draw vertical line through center O, and horizontal line through the center O. Connect O with A (it will be a radius, R). You will have right triangle A,O,crosspoint of AC and horizontal line. You will have triangle with sides 4,1,R. square of R=16+1=17.

@Masterclass_Geometry

2 ай бұрын

great👌

@albaldin

Ай бұрын

No. The sides of the triangle are 4, 1/2, R

@Grizzly01-vr4pn

Ай бұрын

@@Masterclass_Geometry No, not great. 'Wrong' is the word you're grasping for. I'd have hoped for better than that from someone who runs their own geometry channel ☹

@Grizzly01-vr4pn

Ай бұрын

@@albaldin or if you use the same procedure, but 'horizontally' rather than 'vertically', you get a right triangle with legs 3½ and 2, which gives the same hypotenuse (= radius) of (√65)/2.

@albaldin

Ай бұрын

@@Grizzly01-vr4pn Square of ((3/2) + 2) is 25/4 (root 5/2). It's 7/2 and 2

4R^2=2^2+3^2+4^2+6^2R=√65/2

Just draw perpendicular on each of the chords from the centre. Centre = C Perpendicular on chord (7 units) =CP Perpendicular on chord (8 units)=CM Intersecting point of chords =N CPNM a rectangle. CP=MN MN=1/2(6+2)-2=2 CP=2 🔺 CPR (R is the point that intersects circumference and the 7 unit chord (right hand side) Hence CR =√[(7/2)^2+2^2] Comment please

@GeometryDose

12 күн бұрын

Nice

Man, more than solving a problem you proved a theorem!

@GeometryDose

27 күн бұрын

Actually the main purpose I made this video was to prove this theorem. Finding the radius was just a little bonus. Glad you noticed it :).

R=abc/4S

4.03(approximately )

I have shorter and simpler solve.