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प्रश्न
The line 2x − y + 6 = 0 meets the circle x2 + y2 + 10x + 9 = 0 at A and B. Find the equation of circle on AB as diameter.
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उत्तर
2x − y + 6 = 0
∴ y = 2x + 6
Substituting y = 2x + 6 in x2 + y2 + 10x + 9 = 0,
we get
x2 + (2x + 6)2 + 10x + 9 = 0
∴ x2 + 4x2 + 24x + 36 + 10x + 9 = 0
∴ 5x2 + 34x + 45 = 0
∴ 5x2 + 25x + 9x + 45 = 0
∴ (5x + 9) (x + 5) = 0
∴ 5x = – 9 or x = – 5
∴ x = `(-9)/5` or x = – 5
When x = `(-9)/5`
y = `2 xx (-9)/2 + 6`
= `(-18)/5 + 6`
= `(-18 + 30)/5`
= `12/5`
∴ Point of intersection is `"A"((-9)/5, 12/5)`.
When x = – 5,
y = – 10 + 6 = – 4
∴ Point of intersection in B (–5, –4).
By diameter form, equation of circle with AB as diameter is
`(x + 9/5)(x + 5) + (y - 12/5)(y + 4)` = 0
∴ (5x + 9) (x + 5) + (5y – 12) ( y + 4) = 0
∴ 5x2 + 25x + 9x + 45 + 5y2 + 20y – 12y – 48 = 0
∴ 5x2 + 5y2 + 34x + 8y – 3 = 0.
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