Find the equation of a circle which passes through two points
(2, 5) & (5,7). It's centre lies on the line 7x+2y=7.
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Find the equation of the circle which passes through the point (2,−2) and (3,4) and whose centre lies on the line x+y=2.
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We know that the equation of a circle with center at (h,k) is,
(x−h)
2
+(y−k)
2
=r
2
Now, this circle passes through (2,−2). Then
(2−h)
2
+(−2−k)
2
=r
2
h
2
+k
2
−4h+4k+8=r
2
Again, this circle passes through (3,4). Then
(3−h)
2
+(4−k)
2
=r
2
h
2
+k
2
−6h−8k+25=r
2
Subtracting the above results, we have
2h+12k=17 …… (1)
Now, the center lies on x+y=2. So,
h+k=2 ….. (2)
Solving equations (1) and (2), we get
h=
10
7
and k=
10
13
Put values of (h,k) in the first expression.
(2−
10
7
)
2
+(−2−
10
13
)
2
=r
2
r
2
=
100
169+1089
r=
10
1258
Therefore, equation of the circle is,
(x−
10
7
)
2
+(y−
10
13
)
2
=
100
1258
Hence, this is the required equation.
Answer:
7x+2y=7
9xy=7
xy=7-9
=2
answer is this