Question

The area of the circle whose centre is (0, – 4) and passes through the point (–1, 5), is
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Answer 1

Answer:

$r = \sqrt{( {- 1 - 0})^{2} + ( {5 + 4})^{2} } \\ \\ \: = \sqrt{1 + 81} = \sqrt{82} \\ \\ area \: of \: circle = \pi {r}^{2} \\ \\ \: \: \: \: \frac{22}{7} \times 82 = \frac{1804}{7} = 257.7sq.unit$

Answer 2

Answer :

$\green { \boxed{ \sf \: \: Area \: of \: the \: circle \: \Delta = 82\pi}}$

Step-by-step explanation:

Given,

• Center of a circle = (0,-4)
• and it passes through the point (-1,5)

To find :

• Area of the circle.

Solution :

we know that

$\sf \: area \: of \: a \: circle \: \Delta = \pi {r}^{2}$

where r = radius of the circle.

So at first we have to find the radius(r) of the circle.

$\bf \: here \\ \small{\small{ \sf \: radius \: \: r = distance \: between \: center(0, - 4) \: and \: the \: point( - 1,5)}} \\ \sf \implies \: r = \sqrt{ { \{0 - ( - 1)} \}^{2} + {( - 4 - 5})^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{ {(0 + 1)}^{2} + {( - 9)}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{1 + 81} \\ \: \: \: \therefore \: \: \sf \: r\: \: = \sqrt{82} \: unit \\ \\ \bf \: now \: area \: \: \: \Delta = \pi {r}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \pi {( \sqrt{82} })^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 82\pi \\ \\ \green{ \sf \: so \: area \: of \: the \: circle \: \Delta = 82\pi}$