The equation of the circle passing through (3, 0) and (0, 4) and having the minimum radius is
Answers
Answer:
find it on doubt nout I'm sure you will get the correct answer
Step-by-step explanation:
give me thanks ❤️ mark as brainliest follow me
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
Step-by-step explanation:
please mark as brainlist