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प्रश्न
Find the centre and radius of the circle.
x2 + y2 – 8x + 10y – 12 = 0
योग
उत्तर
The equation of the given circle is x2 + y2 – 8x + 10y – 12 = 0.
x2 + y2 – 8x + 10y – 12 = 0
⇒ (x2 – 8x) + (y2 + 10y) = 12
⇒ {x2 – 2(x)(4) + 42} + {y2 + 2(y)(5) + 52} –16 – 25 = 12
⇒ (x – 4)2 + (y + 5)2 = 53
⇒ (x - 4)^2 + {y-(-5)}^2 = (sqrt53)^2`, which is of the form (x – h)2 + (y – k)2 = r2, where h = 4, k = –5, and r = `sqrt53`
Thus, the centre of the given circle is (4, –5), while its radius is `sqrt53`.
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