Example 1 : <\/span>Find the equation of the normal to the circle \\(x^2 + y^2 – 5x + 2y -48\\) = 0 at the point (5,6).<\/p>\n Solution : <\/span>Since the normal to the circle always passes through the centre so the equation of the normal will be the line passing through (5,6) & (\\(5\\over 2\\), -1) Example 2 : <\/span> Find the equation of circle having the lines \\(x^2\\) + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x – 4) + y(y – 3) = 0.<\/p>\n Solution : <\/span>Pair of normals are (x + 2y)(x + 3) = 0 Example 3 : <\/span>The equation of the circle through the points of intersection of \\(x^2 + y^2 – 1\\) = 0, \\(x^2 + y^2 – 2x – 4y + 1\\) = 0 and touching the line x + 2y = 0, is-<\/p>\n Solution : <\/span>Family of circles is \\(x^2 + y^2 – 2x – 4y + 1\\) + \\(\\lambda\\)(\\(x^2 + y^2 – 1\\)) = 0 Practice these given circle examples to test your knowledge on concepts of circles.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some circle examples for better understanding of circle concepts. Example 1 : Find the equation of the normal to the circle \\(x^2 + y^2 – 5x + 2y -48\\) = 0 at the point (5,6). Solution : Since the normal to the circle always passes through the centre so the equation …<\/p>\n
i.e. y + 1 = \\(7\\over {5\/2}\\)(x – \\(5\\over 2\\)) \\(\\implies\\) 5y + 5 = 14x – 35
\n \\(\\implies\\) 14x – 5y – 40 = 0<\/p>\n
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\n \\(\\therefore\\) Normals are x + 2y = 0, x + 3 = 0
\n Point of intersection of normals is the centre of the required circle i.e. \\(C_1\\)(-3,3\/2) and centre of the given circle is \n \\(C_2\\)(2,3\/2) and radius \\(r^2\\) = \\(\\sqrt{4 + {9\\over 4}}\\) = \\(5\\over 2\\)
\n Let \\(r_1\\) be the radius of the required circle
\n \\(\\implies\\) \\(r_1\\) = \\(C_1\\)\\(C_2\\) + \\(r_2\\) = \\(\\sqrt{(-3-2)^2 + ({3\\over 2}- {3\\over 2})^2}\\) + \\(5\\over 2\\) = \\(15\\over 2\\)
Hence equation of required circle is \\(x^2 + y^2 + 6x – 3y – 45\\) = 0\n <\/p>
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(1 + \\(\\lambda\\))\\(x^2\\) + (1 + \\(\\lambda\\))\\(y^2\\) – 2x – 4y + (1 – \\(\\lambda\\))) = 0
\\(x^2 + y^2 – {2\\over {1 + \\lambda}} x – {4\\over {1 + \\lambda}}y + {{1 – \\lambda}\\over {1 + \\lambda}}\\) = 0
Centre is (\\({1\\over {1 + \\lambda}}\\), \\({2\\over {1 + \\lambda}}\\)) and radius = \\(\\sqrt{4 + {\\lambda}^2}\\over |1 + \\lambda|\\)
Since it touches the line x + 2y = 0, hence
Radius = Perpendicular distance from centre to the line
i.e., |\\({1\\over {1 + \\lambda}} + 2{2\\over {1 + \\lambda}}\\over \\sqrt{1^2 + 2^2}\\)| = \\(\\sqrt{4 + {\\lambda}^2}\\over |1 + \\lambda|\\) \\(\\implies\\) \\(\\sqrt{5}\\) = \\(\\sqrt{4 + {\\lambda}^2}\\) \\(\\implies\\) \\(\\lambda\\) = \\(\\pm\\) 1.
\\(\\lambda\\) = -1 cannot be possible in case of circle. So \\(\\lambda\\) = 1.
Hence the equation of the circle is \\(x^2 + y^2 – x – 2y\\) = 0<\/p>\n
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