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Find the Equation of the Circle Whose Centre is (1, 2) and Which Passes Through the Point (4, 6). - Mathematics

Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

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Solution

Let (hk) be the centre of a circle with radius a.
Thus, its equation will be \[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

Given:
h = 1, = 2

∴ Equation of the circle = \[\left( x - 1 \right)^2 + \left( y - 2 \right)^2 = a^2\]

Also, equation (1) passes through (4, 6).

∴\[\left( 4 - 1 \right)^2 + \left( 6 - 2 \right)^2 = a^2\]

\[\Rightarrow 9 + 16 = a^2 \]
\[ \Rightarrow a = 5 \left( \because a > 0 \right)\]

Substituting the value of a in equation (1):

\[\left( x - 1 \right)^2 + \left( y - 2 \right)^2 = 25\]
\[\Rightarrow x^2 + 1 - 2x + y^2 + 4 - 4y = 25\]
\[ \Rightarrow x^2 - 2x + y^2 - 4y = 20\]
\[ \Rightarrow x^2 + y^2 - 2x - 4y - 20 = 0\]

Thus, the required equation of the circle is

\[x^2 + y^2 - 2x - 4y - 20 = 0\]
Concept: Circle - Standard Equation of a Circle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 24 The circle
Exercise 24.1 | Q 3 | Page 21
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