As we know that the standard equation of circle is ${\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}}$ , where r is the radius of circle and centre at (h,k) 

Now , as the circle passes through (2,9) so it must satisfy the above equation after putting the values of h and k respectively

${:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}$

${:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}$

${:\implies \quad \sf 9+16=r^{2}}$

After raising ½ power to both sides , we will get r = +5 , -5 , but as radius can never be -ve . So r = +5 

Now , putting values in our standard equation ;

${:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}$

${:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}$

This is the required equation of Circle

Refer to the attachment as well !

8 0

1 year ago

4x = -36 plz helpppp

san4es73 [151]

Hi there! :)

4x=-36

Divide both sides by 4 in order to isolate x:

x=-9

Hope this helps you and have a nice day!

~Just a joyful teen

$SilentNature$

8 0

1 year ago

Using that formula

d1i1m1o1n [39]

20% x 50 =(20 / 100) x 50 = (20 x 50) / 100 = 1000 / 100 = 10

answer is 10

8 0

2 years ago

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The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If

frutty [35]

For this case we have the following equation:
$w = 0.9* \sqrt{E}$
Where,
w: The weight of a spring in pounds
E: the energy stored by the spring in joules.
Substituting values we have:
$w = 0.9* \sqrt{12}$
Making the corresponding calculation:
$w=3.12$
Answer:
the approximate weight of the spring in pounds is:
$w=3.12$

4 0

2 years ago

Read 2 more answers