5. The Following Note Appears On A Drawing Along With A Leader Pointing To 2 Concentric Circles. O

What is the area of intersection of 2 circles of radius 5.5 cm whose centers are 6 cm apart?

Clearly,You'd forming a figure where you have 2 intersecting congruent circles. The area according to me is 36.2060 centimeter square approximately. Here's a detailed solution to your question-:I've only applied a few basic properties of circles and applied them. Hope my answer does help!

A point is selected randomly from the interior of a circle. The probability that the point is closer to the center than the boundary of circle is?

As we know Probability is the ratio of number of favorable outcomes to the total number of possible outcomes, we have to calculate them first individually.First of all consider the radius of circle as r, then the points closer to center than boundary will lie within the radius of r/2. So the favorable outcomes would be the points inside the area of circle with radius r/2, whereas the total possible outcomes would be all the points inside the area of circle with radius r.Therefore the probability is[math]= \frac{\frac{\pi r^2}{4}}{\pi r^2}[/math][math]= \frac{1}{4}[/math]Although the points on circumference of circle with radius r/2 are equidistant from center and boundary, I think they can be neglected

A line cuts two concentric circles. The length of chords formed by that line on the two circles are 4 cm and 16 cm. What is the difference (in cm²) in squares of radii of two circles?

Let AB be a circle of radius, [math]r[/math] cm and AH be a concentric circle of radius [math]R[/math] cm. A line passes through the circles such that it cuts a chord, [math]CD=4cm[/math] with circle AB and [math]EF=16cm[/math] with circle AH.We need to obtain the difference between the square of the two radii: [math]R^2-r^2[/math]Drop a perpendicular from A to EF i.e, [math]AG\perp EF[/math]. Join EA and EF.We have [math]\triangle AGE\cong\triangle AGF[/math][math]\Rightarrow EG=FG=8cm[/math]Also, in [math]\triangle AGE,AE^2(=R^2)=AG^2+GE^2=AG^2+8^2[/math][math]\Rightarrow AG^2=R^2-64[/math] ...(1)Similarly, join CA and DA.Then [math]\triangle CAG\cong\triangle DAG[/math][math]\Rightarrow CG=DG=2cm[/math]Also, in[math]\triangle CAG,CA^2(=r^2)=AG^2+GC^2=AG^2+2^2[/math][math]\Rightarrow AG^2=r^2-2^2=r^2-4[/math]. …(2)From (1) and (2),[math]R^2-64=r^2-4[/math][math]\Rightarrow R^2-r^2=64-4=60 cm^2[/math]Happy math!!

What is the radius of a circle passing through the point (2,0) & (-4,8)?

Answer: Any real number greater than or equal to 5.The radius of the circle, r, has to be a number greater than or equal to 5. Beyond that, there is absolutely nothing else you can say.Notice that the distance between the two given points is 10 (from the Pythagorean theorem, distance formula, or recognizing 6–8–10 right triangle).Visualize the two points A and B connected by a line segment, AB. So, AB is a chord in the circle. Now draw the perpendicular bisector. If you pick any point P on that new line (the perpendicular bisector of AB), it can serve as a center of the circle, because the distance of A to P matches the distance of B to P (by a characterization property of the perpendicular bisector).Now visualize the point P as very, very close to AB. The distance of P to A and the matching distance of P to B each will be a bit over 5 because the broken line from A to P to B is “almost the same” as segment AB. Segment AB has length 10, so our broken line, which is composed of two radii, has each of AP and BP a bit over 5.Note further that point P may actually be on segment AB. This is the case where r = 5 and the chord is as large as possible, meaning it’s a diameter, with d=2r=10.If you now instead pick P a gazillion miles away, you can see there is no limit to how large the radius can be.

Two chords, KL and MN of a circle intersect at an external point J. If KL = 15 cm. L J = 5cm, MN = 21 cm and NJ = 4 cm, then what will be the ratio of the area of the quadrilateral KLNM to the area of the triangle JKM?

The [math]\triangle{JKM}[/math] and [math]\triangle{JLN}[/math] have a common angle [math]\angle{J}[/math]. This common angle is formed by sides [math]JK[/math] and [math]JM[/math] in [math]\triangle{JKM}[/math], and by sides [math]JL[/math] and [math]JN[/math] in [math]\triangle{JLN}[/math]. We can find the area of the above triangles by the formula [math]A = \frac{1}{2}absin(\theta)[/math] as:[math]A(\triangle{JKM}) = \frac{1}{2}[/math]x[math]JK[/math]x[math]JM[/math]x[math]sin(J)[/math][math]= \frac{1}{2}[/math]x[math](KL + LJ) [/math]x[math](MN + NJ) [/math]x[math]sin(J)[/math][math]= \frac{1}{2}[/math]x[math]20[/math]x[math]25[/math]x[math]sin(J)[/math][math]= 250sin(J)[/math]Similarly, [math]A(\triangle{JLN}) = \frac{1}{2}[/math]x[math]LJ[/math]x[math]NJ[/math]x[math]sin(J)[/math][math]= \frac{1}{2}[/math]x[math]5[/math]x[math]4[/math]x[math]sin(J)[/math][math]= 10sin(J)[/math]So, [math]\frac{A(\square{KLNM})}{A(\triangle{JKM})} = \frac{A(\triangle{JKM}) -A(\triangle{JLN})}{A(\triangle{JKM})}[/math][math]= \frac{250sin(J) - 10sin(J)}{250sin(J)}[/math][math]= 24/25[/math]The ratio of their areas is [math]24:25[/math].

How many common tangents are possible between the two circles [math]x^2+y^2-2x-4y+1=0[/math] and [math]x^2+y^2-12x-16y+91=0[/math] ?

x^2+y^2–2x-4y+1=0(x-1)^2+(y-2)^2=1+4–1=4(x-1)^2+(y-2)^2=(2)^2C1(1,2) and r1=2unitx^2+y^2–12x-16y+91=0(x-6)^2+(y-8)^2=36+64–91=9(x-6)^2+(y-8)^2=(3)^2.C2(6,8) and r2 = 3unit.C1C2=[(6–1)^2+(8–2)^2]^1/2C1C2=(61)^1/2 , r1+r2=2+3=5=(25)^1/Here C1C2 > (r1 + r2)Therefore both the circle neither touch norcut each other. So total four common tangents are possible between the two givencircle. , Answer.

How many circles can be drawn in a circle?

It is definitely NOT infinite. How could it be? Are we talking of imaginary Geometry where such things exists?A point is a circle with ZERO radius, so its area would be ZERO. Which means it does not exists.Now, let’s talk about things that exists. Consider the point as a circle whose radius tends to become ZERO but never EVER becomes ZERO. It is trying hard to become but a strong force of LIMITS is not allowing it. :)Let’s assume the radius of this point is r and r → 0 (tends to become ZERO but never becomes ZERO).So a circle with a radius R (capital R) would have an area of π * R * R.It would be able to accommodate these many points of radius r (r → 0) inside it:(π * R * R) / (π * r * r)And since r is not exactly equal to ZERO, this will never yield INFINITE.Now consider following image.If we assume variables as per the above diagram, then:A circle with radius R would be able to handle only R/W number of inner circles without any separation distance between two consecutive circles.If there should be some separation distance (D) between two consecutive circles, then it would be able to handle R/(W+D).There will be some space left in the centre (centre of outermost circle) which depends on the factor that whether sum of width (W) and distance between two consecutive circles (D) is an exact multiple of radius of outermost circle (R) or not. If it (W + D) is an exact multiple then there will be no space left but if they are not then space left in the centre of the outermost circle would be a circle with a diameter of2(R%(W + D)) (% is modulo operator which returns the remainder after the division)There are more ways where we can draw a circle inside another circle. But the above seems to be the most efficient.Other way is as below. You repeat the small circles until the outermost circle fills up.However, if you like imaginary things, then it is infinite. Live your imaginary life :)