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Equation of the Circle Through Origin Which Cuts Intercepts of Length a and B on Axes is - Mathematics

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प्रश्न

Equation of the circle through origin which cuts intercepts of length a and b on axes is

पर्याय

  •  x2 + y2 + ax + by = 0

  • x2 + y2 − ax − by = 0

  • x2 + y2 + bx + ay = 0

  • none of these

MCQ

उत्तर

x2 + y2 − ax − by = 0

Centre of the circle is \[\left( \frac{a}{2}, \frac{b}{2} \right)\] and its radius is \[\sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2} = \frac{1}{2}\sqrt{a^2 + b^2}\] .

Equation of circle:

\[\left( x - \frac{a}{2} \right)^2 + \left( y - \frac{b}{2} \right)^2 = \frac{1}{4}\left( a^2 + b^2 \right)\]

\[\Rightarrow\] \[\left( 2x - a \right)^2 + \left( 2y - b \right)^2 = \left( a^2 + b^2 \right)\]

\[\Rightarrow\] \[4 x^2 + a^2 - 4ax + 4 y^2 + b^2 - 4by = a^2 + b^2\]

\[\Rightarrow\] \[x^2 - ax + y^2 - by = 0\]

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Circle - Standard Equation of a Circle
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 24: The circle - Exercise 24.6 [पृष्ठ ४०]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 24 The circle
Exercise 24.6 | Q 24 | पृष्ठ ४०

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