The equation of the circle passing through (3,0) and (0,4) and having the minimum radius is
Answers
Answer:
This can be done in two ways. Quick and simple way is:
The circle passes through A(2,0) and B(0,4). There will be infinite number of circles passing through these points. Draw AB. If center lies outside AB, then AB is chord on the circle. Then the diameter of circle is bigger than AB.
If the center of circle is the midpoint of AB and AB is the diameter. So
AB = √(2²+4²) = 2√5 So radius = √5.
Area of the smallest circle = 5 π
===============================
Long way:
(x-a)² + (y-b)² = r²
(2 - a)² + b² = r² => a² - 4 a + 4 + b² = r²
a² + (4-b)² = r² => a² + b² - 8 b +16 = r²
Solving these equations we get a = 2b - 3. Substitute it in the first equation,
b² - 4b + 5 = r²/5 = P(b) = y
d y / d b = 2 b - 4 = 0 => b = 2 . P(b) = y has a minimum value when b = 2.
Minimum value = P(2) = 4 - 8 + 5 = 1
So smallest radius = r = √5
Area of smallest circle = 5π
Answer:
Let (x−h)
2
+(y−k)
2
=r
2
…………..(1)
where (h, k) is the centre of the circle and r is the radius.
Let us draw a line A(2,0) and B(4,0). If the centre lies outside the line AB then it is a chord.
The diameter of the circle is bigger than the chord. Suppose AB is the diameter of the circle then the centre of the circle must be a midpoint of AB.
∴ Using section formula
h=
2
2+0
=
2
2
=1
k=
2
4+0
=
2
4
=2
∴ Coordinates of the centre=(h,k)
=(1,2)
Equation (1) becomes,
(x−1)
2
+(y−2)
2
=r
2
…………….(2)
Since (2) passes through (2,0), equation (2) can be written as,
(2−1)
2
+(0−2)
2
=r
2
1
2
+2
2
=r
2
1+4=r
2
r
2
=5
r=
5
Equation of the circle with minimum radius is
∴(x−1)
2
+(y−2)
2
=5
⇒x
2
+1−2x+y
2
+4−4y=5
⇒x
2
+y
2
−2x−4y+5−5=0
⇒x
2
+y
2
−2x−4y=0.
solution