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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

ધારો કે વર્તુળનું સમીકરણ, જે \(x\)-અક્ષને બિંદુ \((a, 0), a\gt0\) પર સ્પર્શે છે અને \(y\)-અક્ષ પર \(b\) લંબાઈનો આંતરછેદ કાપે છે, તે \(x^2+y^2-\alpha x+\beta y+\gamma=0\) છે. જો વર્તુળ \(x\)-અક્ષની નીચે આવેલું હોય, તો ક્રમયુક્ત જોડ \(\left(2 a, b^2\right)\) = __________

  1. A \(\left(\gamma, \beta^2-4 \alpha\right)\)
  2. B \(\left(\alpha, \beta^2+4 \gamma\right)\)
  3. C \(\left(\gamma, \beta^2+4 \alpha\right)\)
  4. D \(\left(\alpha, \beta^2-4 \gamma\right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\left(\alpha, \beta^2-4 \gamma\right)\)

Step-by-step Solution

Detailed explanation

By pythagoras \(\mathrm{r}^2=\mathrm{a}^2+\frac{\mathrm{b}^2}{4}=\mathrm{P}^2\) \(r=\sqrt{\frac{4 a^2+b^2}{4}}\) Equation of circle is \((x-\alpha)^2+(y-\beta)^2=r^2\) \(x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0\) comparision \(x^2+y^2-\alpha x+\beta y+r=0\)…
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