qk Prove that the distances of two points from the centre of a circle are proportional to the distances of each from the polar of the other. a
Answers
Given two points (x1,y1) and (x2,y2), recall that their horizontal distance from one another is Δx=x2−x1 and their vertical distance from one another is Δy=y2−y1. (Actually, the word "distance'' normally denotes "positive distance''. Δx and Δy are signed distances, but this is clear from context.) The actual (positive) distance from one point to the other is the length of the hypotenuse of a right triangle with legs |Δx| and |Δy|, as shown in figure 1.2.1. The Pythagorean theorem then says that the distance between the two points is the square root of the sum of the squares of the horizontal and vertical sides:
distance=(Δx)2+(Δy)2−−−−−−−−−−−−√=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√.
For example, the distance between points A(2,1) and B(3,3) is (3−2)2+(3−1)2−−−−−−−−−−−−−−−√=5–√.
Condition for cutting at right angle'
(distancebetweencentre)
2
=(radius)
1
2
+(radius)
2
2
1st circle
r=acos(θ−α)
x=rcosθ
y=rsinθ
r=acosθcosα+asinθsinα
r=
r
ax
cosα+
r
ay
sinα
x
2
+y
2
=axcosα+aysinα
(x−
2
a
cosα)
2
+(y−
2
a
sinα)
2
=(
2
a
)
2
2nd circle
r=asin(θ−α)
r=asinθcosα−asinαcosθ
x=rcosθ
y=rsinθ
r=
r
ay
cosα−
r
ax
sinα
r
2
=aycosα−axsinα
x
2
+y
2
=aycosα−axsinα
(x+
2
a
sinα)
2
+(y−
2
a
cosα)
2
=(
2
a
)
2
Centre of two circle
(
2
a
cosα,
2
a
sinα),(
2
−a
sinα,
2
a
cosα)
and radius of both =
2
a
Distance between centers =(
2
a
cosα+
2
a
sinα)
2
+(
2
a
sinα−
2
a
cosα)
2
=
4
a
2
+
4
a
2
=
2
a
2
Hence proved, the circles intersect at right angles.