Question

Ques no:(11)
Ml_aggarwal_APC (Class11) bk
chapter= Circle​
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Q: Find the equation of circle
i) Which touches the x-axis on the positive direction at a distance 5 units from the origin and have radius 6 units.
ii) passing through the origin, radius 17 and ordinate of the centre is - 15

iii) which touches both the axes and pass through the point (2,1).
.......... ......... ........ ....
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Answer 1

Answer:

$hello$

(i)"

which touches both the axes at a distance of 6 units from the origin.

A circle touches the axes at the points (±6, 0) and (0, ±6).

So, a circle has a centre (±6, ±6) and passes through the point (0, 6).

We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

So, the equation is (x – p)2 + (y – q)2 = r2

Where, p = 6, q = 6

(x – 6)2 + (y – 6)2 = r2 ….

(1) Equation (1) passes through (0, 6)

So, (0 – 6)2 + (6 – 6)2 = r2..

36 + 0 = r2

r = √36

= 6

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2..

Now by substituting the values in the equation, we get (x ± 6)^2 + (y ± 6)^2 = (6)^2

x2 ± 12x + 36 + y2 ± 12y + 36 = 36

x2 + y2 ± 12x ± 12y + 36 = 0

∴ The equation of the circle is x2 + y2 ± 12x ± 12y + 36 = 0...

$< hope \: it \: will \: help \: uh... >$

Answer 2

Answer:

hope it helps you. I have answers to the whole chapter.

Step-by-step explanation:

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