问题11：半径为4个单位的圆接触第一象限中的坐标轴。求出相对于线镜x = 0和y = 0的图像方程式
解决方案：
Centre of the circle is (4, 4)
Equation of the circle is (x-4)2 + (y-4)2 =16
Images of this circle with respect to the line mirrors x=0 and y=0. They have centres at (-4, 4) and (4, -4) respectively.
Required equations are (x+4)2 + (y-4)2 =16 and (x-4)2 + (y+4)2 =16
⇒x2 + y2 +8x -8y +16=0 and x2+y2-8x+8y+16=0
问题12.找到在(0，3)处与y轴接触的圆的方程，并在x轴上截取8个单位。
解决方案：
Centre of the considered circle is (1, 1)
Radius of the considered circle is 1
The circle is rolled along the positive direction of the x-axis. On completing one roll, its centre moves horizontally through a distance equal to its circumference, i.e 2π.
Therefore, the coordinates of the centre of the new circle will be (1+2π, 1).
The required equation of the circle is (x-1-2π)2 +(y-1)2=1
问题13：找到通过Y轴上两个点的圆的方程，这些圆与原点的距离为3，半径为5。
解决方案：
The circle passes through the points(0, 3) and (0, -3).
So, (0-h)2 +(3-k)2 = a2 …(1) and (0-h)2 +(-3 -k)2=a2 …(2)
Solving (1) and (2), k=0
Given, radius is 5 units.
Therefore, a2 = 25
From equation (2) we get, h2 +9 =25 ⇒ h=±4
So, the required equation is (x±4)2 +y2 = 25 ⇒ x2 ± 8x +y2 -9 = 0
问题14.如果线2x -3y = 5和3x – 4y = 7是面积为154平方单位的圆的直径，则该圆的方程式。
解决方案：
πr2 = 154 ⇒ r2 = 49
The point of intersection of both the lines will be the centre of the circle.
Solving 2x -3y = 5 and 3x – 4y = 7 we get, x=1 and y=-1
Putting the values in the standard equation, we get (x-1)2 +(y+1)2 =49
⇒ x2 + y2 – 2x + 2y -47 = 0
问题15。如果线y =√3x + k接触圆x 2 + y 2 = 16，则求出k的值。
解决方案：
From the given equation of circle we can conclude the centre is at (0, 0) and radius is 4.
The perpendicular distance from the centre of the circle to the tangent y = √3 x + k is equal to the radius of the circle.
Using the formula of perpendicular distance of point from a line, we get k=±8.
问题16。找到以(1，-2)为中心并经过线3x + y = 14和2x + 5y = 18的交点的圆的方程
解决方案：
Solving 3x+y=14 and 2x+5y=18 we get
x=4 and y=2
The radius is equal to the distance between (1,-2) and (4,2)
r=√((4-1)2+(2+2)2)
=√(9+16)
=5
Putting values in the standard equation,
⇒(x-1)2+(y+2)2=25
⇒x2+y2-2x+4y-20=0
问题17.如果线3x-4y + 4 = 0和6x-8y-7 = 0与圆相切，则找到圆的半径。
解决方案：
Given, 3x-4y+4=0 and 6x-8y-7=0
⇒y=3/4x+1 and y=3/4x-7/8
The slope of both the lines is equal so, both the lines are parallel.
Using the formula of the distance between the parallel lines we get, 3/2
The radius is equal to the half of the distance between the parallel lines (diameter of the circle).
Therefore , the radius is given by 3/4.
问题18显示的是(X，Y)用x = 2原子/ 1 + T 2中给出的点和Y = A(1-T 2/1 + T 2)位于一个圆上为T，使得所有实数值- 1 <= t <= 1，其中a是任何给定的实数。
解决方案：
Square and add x=2at/1+t2 and y=a(1-t2/1+t2), we get
x2+y2=(2at/1+t2)2+a2(1-t2/1+t2)2
⇒x2+y2=(4a2t2+a2-2a2t2+a2t2)/(1+t2)2
⇒x2+y2=a2((1+t2)2/((1+t2)2))
⇒x2+y2=a2
The above equation represents the equation of a circle, hence points (x, y) lie on the circle.
问题19.圆x 2 + y 2 -2x-2y + 1 = 0沿x轴的正方向滚动并完成一整圈。在新位置找到其方程式。
解决方案：
Centre of the circle is (1,1)
Radius of the circle is 1
The circle is rolled along the positive direction of the x-axis. On making one complete roll, its centre moves horizontally a distance equal to its circumference i.e. 2π
So, the coordinate of the centre of the new circle will be (1+2π,1)
Therefore, the required equation of the circle is (x-1-2π)2 +(y-1)2=1
问题20.外接矩形ABCD的圆的一个直径为4y = x + 7。如果A和B的坐标分别为(-3,4)和(5,4)，请找到圆的方程。
解决方案：
The centre of the circle lies on the line 4y=x+7 and circle passes through A(-3,4) and B(5,4).
Slope of the segment joining A and B is zero .
So, the slope of the perpendicular bisector of AB is not defined.
The perpendicular bisector of AB will be parallel to the y-axis and pass through ((-3+5/2), (4+4/2)) = (1,4)
Equation of the perpendicular bisector is x =1
Intersection point of the perpendicular bisector and 4y=x+7 is (1,2)
Therefore, centre=(1,2) and Radius=√((5-1)2+(4-2)2)=√20
The required equation of the circle is x2+y2-2x-4y-15=0
问题21.如果线2x-y + 1 = 0接触点(2，5)处的圆，并且圆的中心位于线x + y-9 = 0上。找出圆的方程。
解决方案：
Let coordinates of the centre be (t, 9-t) and the radius be a.
Using the formula of perpendicular distance of line from a point, we get
a = |(3t-8)/√5| ⇒ a2 =((3t-8)/√5)2 …(1)
From the deductions, the equation of the circle is (x-t)2+(y-(9-t))2=a2 …(2)
The circle passes through (2, 5). Substituting values we get, t=6
Substituting t in (1), a2 =(10/√5)2
Putting the values found in (2), the required equation is,
(x-6)2+(y-3)2=20