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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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