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  • What is the equation of a circle with a centre (6,-2) and passing through (2,-2)?Option: 1 <img alt="\left ( x-6 \right )^{2}+\left ( y+2 \

What is the equation of a circle with a centre (6,-2) and passing through (2,-2)?

Option: 1

\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 16


Option: 2

\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 36


Option: 3

\left ( x-2 \right )^{2}+\left ( y+2 \right )^{2}= 16


Option: 4

\left ( x-2 \right )^{2}+\left ( y+2 \right )^{2}= 36


Answers (1)

best_answer

The equation of a circle with centre (h,k) and radius r is given by \left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

We are given that the centre of the circle is (6,-2) and it passes through (2,-2).

Here,
h= 6
k= -2

Using the distance formula, we can find the radius of the circle:

r= distance\: between\left ( 6,-2 \right )and\left ( 2,-2 \right )
= \sqrt{\left ( 6-2 \right )^{2}+\left ( -2-2\left ( -2 \right ) \right )^{2}}
= \sqrt{\left ( 16+0 \right )}
= 4

Therefore ,r= 4

Substituting the values of (h,k,r) into the equation of a circle, we get:
\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 4^{2}
\left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 16

So, the correct answer is the option (a) \left ( x-6 \right )^{2}+\left ( y+2 \right )^{2}= 16

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Pankaj

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