SOLUTION: A circle through P(5,9), Q(11,-3), and R(13,3). Find its equation in standard form.
The perpendicular bisector of any chord of a circle will pass through the center. Thus, find th
Algebra.Com
Question 1183738: A circle through P(5,9), Q(11,-3), and R(13,3). Find its equation in standard form.
The perpendicular bisector of any chord of a circle will pass through the center. Thus, find the intersection of the two perpendicular bisectors above. This will be the center C of the circle.
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
How many times to ask the same exact question?
RELATED QUESTIONS
A circle through P(5,9), Q(11,-3), and R(13,3). Find its equation in standard form.... (answered by josgarithmetic)
A circle through P(5,9), Q(11,-3), and R(13,3). Find its equation in standard form, by... (answered by greenestamps)
A circle through P(5,9), Q(11,-3), and R(13,3). Find its equation in standard form.
What (answered by josgarithmetic)
A circle through P(5,9), Q(11,-3), and R(13,3). Find its equation in standard form.
To... (answered by josgarithmetic)
In advance I apologize for this long question and also thank you for helping me.
Use... (answered by josgarithmetic)
T(-10,-2) R(10,8) Y(11,-9) are vertices of ∆TRY
1) find equation of YA, the... (answered by mananth)
The radius of a circle is 5 and its center is at (-3,-4). Find the lenght of the chord... (answered by nerdybill)
If P is a point on the circle x2+y2=9 and Q is a point on the line 7x+y+3=0 and the line... (answered by KMST)
If P is a point on the circle x2+y2=9,Q is a point on the line 7x+y+3=0 and the line... (answered by KMST)