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Find the Equation of the Circle Which Has Its Centre at the Point (3, 4) and Touches the Straight Line 5x + 12y − 1 = 0. - Mathematics

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प्रश्न

Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y − 1 = 0.

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उत्तर

It is given that the centre is at the point (3, 4).
Let the equation of the circle be

\[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

∴ Equation of the required circle =

\[\left( x - 3 \right)^2 + \left( y - 4 \right)^2 = a^2\]...(1)
Also, the circle touches the straight line 5x + 12y − 1 = 0.
\[\therefore a = \left| \frac{5\left( 3 \right) + 12\left( 4 \right) - 1}{\sqrt{5^2 + {12}^2}} \right| = \left| \frac{62}{13} \right|\]
\[ \Rightarrow a^2 = \left| \frac{5\left( 3 \right) + 12\left( 4 \right) - 1}{13} \right| = \frac{3844}{169}\]
So, from equation (1), we have:
\[\left( x - 3 \right)^2 + \left( y - 4 \right)^2 = \frac{3844}{169}\]
\[\Rightarrow x^2 + y^2 - 6x - 8y = \frac{3844}{169} - 25\]
\[\Rightarrow 169\left( x^2 + y^2 - 6x - 8y \right) + 381 = 0\]
Hence, the required equation of the circle is
\[169\left( x^2 + y^2 - 6x - 8y \right) + 381 = 0\]
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Advanced Concept of Circle - Standard Equation of a Circle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 24: The circle - Exercise 24.1 [पृष्ठ २१]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 24 The circle
Exercise 24.1 | Q 8 | पृष्ठ २१

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