Question

3. Prove that the circle x2 + y2 - 6x - 2y + 9 = 0
(i) touches the x-axis;
(ii) lies entirely inside the circle x2 + y2 = 18.​

Answer 1

SOLUTION

TO PROVE

The circle x² + y² - 6x - 2y + 9 = 0

(i) touches the x-axis

(ii) lies entirely inside the circle x² + y² = 18.

EVALUATION

(i) Here the given equation of the first circle is

x² + y² - 6x - 2y + 9 = 0

Which can be rewritten as

$\sf{ {(x - 3)}^{2} + {(y - 1)}^{2} = 1}$

So the centre is (3,1) & radius = 1 unit

Since ordinate of the centre = numerical value of radius

Hence the circle touches the x-axis

(ii) The equation of the second circle is

x² + y² = 18

Which can be rewritten as

$\sf{ {(x - 0)}^{2} + {(y - 0)}^{2} = {(3 \sqrt{2} )}^{2} }$

Hence the centre of the circle is (0,0) and radius = 3√2 unit

Hence the circle lies entirely inside the circle x² + y² = 18

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