# Find the Equation of the Circle Whose Centre Lies on the Positive Direction of Y - Axis at a Distance 6 from the Origin and Whose Radius is 4. - Mathematics

Find the equation of the circle whose centre lies on the positive direction of - axis at a distance 6 from the origin and whose radius is 4.

#### Solution

Let (hk) be the centre of a circle with radius a.
Thus, its equation will be

$\left( x - h \right)^2 + \left( y - k \right)^2 = a^2$
The centre of the required circle lies on the positive direction of the y-axis at a distance 6 from the origin.
Thus, the coordinates of the centre are (0, 6).
∴ h = 0, = 6
∴ Equation of the circle = $\left( x - 0 \right)^2 + \left( y - 6 \right)^2 = a^2$
Also, a = 4
Substituting the value of a in equation (1): $\left( x - 0 \right)^2 + \left( y - 6 \right)^2 = 16$
$\Rightarrow x^2 + y^2 + 36 - 12y = 16$
$\Rightarrow x^2 + y^2 - 12y + 20 = 0$
Hence, the required equation of the circle is
$x^2 + y^2 - 12y + 20 = 0$
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 5 | Page 21
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