equation of a circle passing through the point 1 2 and 3 4 and touching the line 3 x + y - 3 equals to zero is
Answers
Answer:
Circle (1,2) and (3,4)−3x+y−3=0⇒(0,1)(1,0)
Let (x,y) be centre of circle ⇒ Radius is equal both points.
(x+1)
2
+(y−2)
2
=
(x−3)
2
(y−2)
2
....(1)
x
2
+1−2x+y
2
+4−4y=x
2
+9−6x+y
2
+16−8y
−20+4x+4y=0⇒x+y=5⇒y=5−x......(2)
distance from center of circle is same as radius.
(3)
2
+(1)
2
∣3x+y−3∣
=
10
∣3x+5−x−3∣
=
10
∣2x+2∣
....(3)
Substituting (2) in (1)4 equality to 3
(x−1)
2
+(3−x)
2
=
10
∣2x+2∣
⇒(x−1)
2
+(3−x)
2
=
10
(2x+2)
2
x
2
+1−2x+9+x
2
−6x=4x
2
+4
10
x
16x
2
−88x+96=0
Solving form =
2a
−b±
b
2
−4ac
=
2×16
88+±
(88)
2
−4×16×96
=
32
88±40
x
1
=
32
86+40
,x
2
=
32
88−40
x
1
=4,x
2
=1.5
Substituting x
1
and x
2
in (2) y
1
=1,y
2
=3.5
we get two circle with center (4,1) and (1.5,3.5)
Radius of circle C
1
=(4−1)
2
+(3−4)
2
=
10
C
2
=(1.5−1)
2
+(3−1.5)
2
=
2.5
C
1
⇒(x−y)
2
+(y−1)
2
=10C
2
⇒(x−1.5)
2
+(y−3.5)
2
=2.5
Step-by-step explanation:
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