Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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# The Equation of the Circle Passing Through the Origin Which Cuts off Intercept of Length 6 and 8 from the Axes is - Mathematics

MCQ

The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is

#### Options

• x2 + y2 − 12x − 16y = 0

• x2 + y2 + 12x + 16y = 0

• x2 + y2 + 6x + 8y = 0

• x2 + y2 − 6x − 8y = 0

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#### Solution

x2 + y2 − 6x − 8y = 0

The centre of the required circle is $\left( \frac{6}{2}, \frac{8}{2} \right) = \left( 3, 4 \right)$ .

The radius of the required circle is

$\sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
Hence, the equation of the circle is as follows:
$\left( x - 3 \right)^2 + \left( y - 4 \right)^2 = 5^2$
$\Rightarrow$ $x^2 + y^2 - 6x - 8y = 0$
Concept: Circle - Standard Equation of a Circle
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 24 The circle
Q 15 | Page 40
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