Question
Find the center and radius of the circle
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Text solutionVerified
$$\displaystyle x^{2}+y^{2}-4x - 8y - 45 = 0$$
$$\displaystyle \Rightarrow \left ( x^{2}- 4x \right ) + \left ( y^{2} - 8y\right )=45$$
$$\displaystyle
\Rightarrow $$ $$\displaystyle \left \{ x^{2}-2\left ( x \right )\left
( 2 \right )+2^{2} \right \}+\left \{ y^{2} - 2\left ( y \right )\left (
4 \right )+4^{2} \right \}- 4 - 16 = 45$$
$$\displaystyle \Rightarrow \left ( x - 2 \right )^{2}+ \left ( y- 4 \right )^{2} = 65$$
$$\displaystyle
\Rightarrow $$ $$\displaystyle \left ( x - 2 \right )^{2}+\left ( y - 4
\right )^{2}=\left ( \sqrt{65} \right )^{2}$$
Which is of the form $$\displaystyle \left ( x - h \right )^{2}+\left ( y - k^{2} \right
)= r^{2}$$
where $$h = 2, k = 4$$ and $$r = \displaystyle \sqrt{65}$$
Thus the centre of the given circle is $$(2, 4)$$ while its radius is $$\displaystyle \sqrt{65}$$
$$\displaystyle \Rightarrow \left ( x^{2}- 4x \right ) + \left ( y^{2} - 8y\right )=45$$
$$\displaystyle
\Rightarrow $$ $$\displaystyle \left \{ x^{2}-2\left ( x \right )\left
( 2 \right )+2^{2} \right \}+\left \{ y^{2} - 2\left ( y \right )\left (
4 \right )+4^{2} \right \}- 4 - 16 = 45$$
$$\displaystyle \Rightarrow \left ( x - 2 \right )^{2}+ \left ( y- 4 \right )^{2} = 65$$
$$\displaystyle
\Rightarrow $$ $$\displaystyle \left ( x - 2 \right )^{2}+\left ( y - 4
\right )^{2}=\left ( \sqrt{65} \right )^{2}$$
Which is of the form $$\displaystyle \left ( x - h \right )^{2}+\left ( y - k^{2} \right
)= r^{2}$$
where $$h = 2, k = 4$$ and $$r = \displaystyle \sqrt{65}$$
Thus the centre of the given circle is $$(2, 4)$$ while its radius is $$\displaystyle \sqrt{65}$$
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Question Text | Find the center and radius of the circle |
Answer Type | Text solution:1 |
Upvotes | 150 |