Nightshade7456
New member
- Joined
- Oct 13, 2020
- Messages
- 6
Please show us what you have tried and exactly where you are stuck.
\((x-h)^2+(y-k)^2=R^2\) is a circle that is centred at \((h,k)\) that has radius \(R>0\).
I need to find answer for c part??????????\((x-h)^2+(y-k)^2=R^2\) is a circle that is centred at \((h,k)\) that has radius \(R>0\).
Thus with this problem we have
\(x^2+y^2+16y+k=0\\(x-0)^2+(y+8)^2-64+k=0\\x^2+(y+8)^2=64-k\)
Now we are told that the radius is \(10\) , so what?
Please consider "Completing the Square" for the y-portions.
Did you calculate the value of k?I need to find answer for c part??????????
The table's been set up for you.I need to find answer for c part??????????\((x-h)^2+(y-k)^2=R^2\) is a circle that is centred at \((h,k)\) that has radius \(R>0\).
Thus with this problem we have
\(x^2+y^2+16y+k=0\\(x-0)^2+(y+8)^2-64+k=0\\x^2+(y+8)^2=64-k\)
Now we are told that the radius is \(10\) , so what?
a spoon!Beer soaked comment/query follows.
The table's been set up for you.
Food is on the table.
What're you waiting for?
k=54Did you calculate the value of k?
Incorrect!k=54
Incorrect as noted in post #12.Here the wrkings
tnx got itBeer soaked hint follows.
Incorrect as noted in post #12.
Requirement (a) was already worked out for you by pka at post #5.
For requirement (b), you are asked to solve for k.
The problem states that the radius is 10.
You should first determine what the value of k is by equating 64 - k with \(\displaystyle 10^2\), or solve for k in \(\displaystyle 64 - k = 10^2\)
After you do that, more hints will come your way.
I don't think that a spoon is enough. Possibly the OP wants to be spoon fed?a spoon!