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the equation of the circle having centre on the line x+2y=3 and passing through the point of intersection of x2+y2-2x-4y+1=0 and x2+y2-4x-2y+4=0
the equation of the circle having centre on the line x+2y=3 and passing through the point of intersection of x2+y2-2x-4y+1=0 and x2+y2-4x-2y+4=0

```
6 years ago

bharat bajaj
IIT Delhi
122 Points
```							The equation of the curve passing through the point of intersection will be :x2+y2-2x-4y+1 + A(x2+y2-4x-2y+4) = 0A is some parameterx2(A+1) + y2(A+1) + x(-2-4A) +y(-4 - 2A) +1 + 4A = 0The centre of this circle :( 1+2A/A+1 , A + 2/A+1)It passes through x+ 2y = 3Hence, 1 + 2A + 2A + 4 = 3A + 35 + 4A = 3A + 3A = -2The equation of circle is = x2 + y2 - 6x + 7 = 0ThanksBharat BajajIIT Delhiaskiitians faculty
```
6 years ago
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• 731 Video Lectures
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• Test paper with Video Solution
• Mind Map
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• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions