# Find the Equation of a Circle Which Touches Both the Axes and Passes Through the Point (2, 1). - Mathematics

Find the equation of a circle
which touches both the axes and passes through the point (2, 1).

#### Solution

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

Let the required equation of the circle be

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

It is given that the circle touches both the axes.
Thus, the required equation will be

$x^2 + y^2 - 2ax - 2ay + a^2 = 0$

Also, the circle passes through the point (2, 1).
∴ $4 + 1 - 4a - 2a + a^2 = 0$

$\Rightarrow a^2 - 6a + 5 = 0$
$\Rightarrow a^2 - 5a - a + 5 = 0$
$\Rightarrow a = 1, 5$

Hence, the required equation is $x^2 + y^2 - 2x - 2y + 1 = 0$ or

$x^2 + y^2 - 10x - 10y + 25 = 0$
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 7.3 | Page 21