14. If the straight line ar+ 3y = 1 intersects the circle +4 at the points A and B, then
find equation of the circle having, AR as diameter.
Answers
To find the intersection points of circle and the line we need to solve their equations simultaneously.
∵2x+3y=1⇒3y=1−2x⇒y=
3
1−2x
Putting this value of y in equation of circle, x
2
+(
3
1−2x
)
2
=4
⇒9x
2
+1+4x
2
−4x=36⇒13x
2
−4x−35=0
∴x=
2×13
−(−4)±
(−4)
2
−4×13×(−35)
=
26
4±6
51
=
13
2±3
51
∴y=
3
1−2[
13
2±3
51
]
⇒y=
13
3∓2
51
Thus the coordinates of A and B are as follows:- A≡(
13
2+3
51
,
13
3−2
51
) and B≡(
13
2−3
51
,
13
3+2
51
)
Now, equation of circle with ends of diameter given is:-(x−x
1
)(x−x
2
)+(y−y
1
)(y−y
2
)=0
⇒(x−
13
2+3
51
)(x−
13
2−3
51
)+(y−
13
3−2
51
)(y−
13
3+2
51
)=0
⇒x
2
−[
13
2+3
51
+
13
2−3
51
]x+(
13
2+3
51
)(
13
2−3
51
)+
y
2
−[
13
3−2
51
+
13
3+2
51
]y+(
13
3−2
51
)(
13
3+2
51
)=0
⇒x
2
−
13
4
x+(
169
4−9×51
)+y
2
−
13
6
y+(
169
9−4×51
)=0
⇒x
2
+y
2
−
13
4
x−
13
6
y+(
169
4+9−(9+4)×51
)=0
⇒x
2
+y
2
−
13
4
x−
13
6
y−
13
50
=0
Thus, c=
13
−50