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Find the Equation of a Circle Passing Through Points P(4,10), Q(6,2), R(12,2)
Mathematics
Grade 10 (Junior High School)
Question Content
The point R has coordinates (12,2). A circle passes through the points P, Q and R. The chord QR is horizontal. (b) Find the equation of the circle. The given points are P(4,10), Q(6,2), R(12,2).
Correct Answer
(x-9)^2+(y-6)^2=25
Detailed Solution Steps
1
Step 1: Recall that the general equation of a circle is (x-a)^2+(y-b)^2=r^2, where (a,b) is the center and r is the radius. We need to find a, b, r using the three points on the circle.
2
Step 2: Substitute point Q(6,2) into the equation: (6-a)^2+(2-b)^2=r^2
3
Step 3: Substitute point R(12,2) into the equation: (12-a)^2+(2-b)^2=r^2
4
Step 4: Set the two equations from Step 2 and 3 equal to each other since both equal r^2: (6-a)^2+(2-b)^2=(12-a)^2+(2-b)^2. Cancel out (2-b)^2 from both sides, expand the squares: 36-12a+a^2=144-24a+a^2. Simplify to get 12a=108, so a=9.
5
Step 5: Substitute point P(4,10) and a=9 into the circle equation: (4-9)^2+(10-b)^2=r^2, which simplifies to 25+(10-b)^2=r^2.
6
Step 6: Substitute point Q(6,2) and a=9 into the circle equation: (6-9)^2+(2-b)^2=r^2, which simplifies to 9+(2-b)^2=r^2.
7
Step 7: Set the equations from Step 5 and 6 equal: 25+(10-b)^2=9+(2-b)^2. Expand the squares: 25+100-20b+b^2=9+4-4b+b^2. Simplify to get 125-20b=13-4b, so 16b=112, b=6.
8
Step 8: Substitute a=9 and b=6 into the equation from Step 6: 9+(2-6)^2=r^2, so 9+16=r^2, r^2=25.
9
Step 9: Write the final equation of the circle: (x-9)^2+(y-6)^2=25
Knowledge Points Involved
1
General Equation of a Circle
The standard form is (x-a)^2+(y-b)^2=r^2, where (a,b) is the center of the circle and r is the radius. This equation is used to define all points (x,y) that lie on the circumference of the circle, which are all equidistant from the center (a,b) by distance r.
2
Perpendicular Bisector of a Chord
The perpendicular bisector of any chord of a circle passes through the center of the circle. For the horizontal chord QR, its perpendicular bisector is a vertical line x=9 (the midpoint of QR is ((6+12)/2, (2+2)/2)=(9,2), so the perpendicular bisector is the vertical line through this point), which gives us the x-coordinate of the center directly.
3
Solving Systems of Equations
When we have multiple unknowns (a,b,r here), we substitute the coordinates of points on the circle into the circle equation to create a system of equations, then solve the system by equating or eliminating terms to find the values of the unknowns.
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