Circle is a closed shape plane figure whose boundary (the circumference) consists
of points equidistant from a fixed point (the centre).
Circle may also be defined as a locus of a point which moves so that its distance from a fixed point is constant. The fixed point is called the centre and the constant distance is called the radius of the circle.
The circle has been known since before the beginning of recorded history. Natural circles would have been observed, such as the Moon, Sun, and a short plant stalk blowing in the wind on sand, which forms a circle shape in the sand. The circle is the basis for the wheel, which, with related inventions such asgears, makes much of modern machinery possible. In mathematics, the study of the circle has helped inspire the development of geometry, astronomy, and calculus.
Some of the elements in the circle are given in alongside figure.
If r be the radius of the circle from the origin and (x,y) be any coordinates on circumference of the circle then.
$$x^2+y^2=0$$
If (h,k) be the center of circle of radius r and (x,y) be any point on the circumference, then.
$$(x-h)^2+(y-k)^2=0$$
Consider an equation
$$x^2+y^2+2gx+2fy+c+0.......(i)$$
$$or,(x^2+2gx+g^2)+(y^2+2fy+f^2)=g^2+f^2+c$$
$$or,(x+g)^2+(y+f)^2=g^2+f^2+c$$
$$or,[x-(-g)]^2+[y-(-f)]^2=\sqrt{(g^2+f^2-c)^2}$$
Comparing the above equation with
$$(x-h)^2+(y-k)^2=r^2$$
Now we get
$$h=-g$$
$$k=-f$$
$$r=\sqrt{g^2+f^2-c}$$
Hence eqution (i) represent an quation of circle with:-
$$centre=(-g,-f)$$
$$radius=\sqrt{g^2+f^2-c}$$
As usual let (h,k) be center of circle and r be radius of it in all cases . Now
(i) Equation of circle touching x-axis
$$(x-h)^2+(y-k)^2=k^2$$
This is because if the circle touches x-axis then the radius of circle(r)=k
(ii) Equation of circle touching y-axis
$$(x-h)^2+(y-k)^2=h^2$$
This is because if the circle touches y-axis then the radius of circle(r)=h
(iii) Equation of circle touching both axes
$$(x-h)^2+(y-h)^2=h^2$$
This is because if the circle touches both axes then the radius of circle(r)=h=k
(iv) Equation of circle in diameter form
Let the co-ordinates of diameter is as shown in figure .
Also, construct a straight line from both A and B meeting at the same point P on the circumference .
Now we can see right angle triangle APB where slopes of
$$(i)AP=\frac{y-y_1}{x-x_1}$$
$$(ii)BP=\frac{y-y_2}{x-x_2}$$
Being AP and BP perpendicular so their product should be -1.
$$or,\frac{y-y_1}{x-x_1}.\frac{y-y_2}{x-x_2}=-1$$
$$(x-x_1)(x-x_2)=(y-y_1)(y-y_2)$$
It is the required equation of circle in diameter form.
1.Find the equation of circle which touches the axes at (2,0) and (0,2)?
2.Find the equation of circle whose center is at (3,4) and touches the x-axis
3.Find the center and radius of the circle x^{2}+y^{2}+4x-6y+4=0?
4.Proove that the circles x+y+2ax+c=0 and x+y+2by+c=0 touch each other if
$$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$$
5.Find the equation of circle which touches the coordinate axes at (a,0) and (0,a)?
6.Find the equation of tangent to the circle x^{2}+y^{2}=a^{2}the point (x_{1},y_{1}) ?
7.Find the equation of circle with two of the diameter are x+y=6 and x+2y+8 and radius=10?
8.If one end of the diameter of the circle x^{2}+y^{2}-2x+6y-15=0 is (4,1), find the coordinate of the end?
Some formulae or keypoints are:
General equation of circle
$$x^2+y^2+2gx+2fy+c=0$$
Whose radius is
$$\sqrt{g^2+f^2-c}$$
And centre
$$(-g,-f)$$
Standard form of circle is
$$x^2+y^2=r^2$$
Standard form of circle with center at any point
$$(x-h)^2+(y-k)^2=r^2$$
Equation of circle in diameter form
$$(x-x_1)(x-x_2)=(y-y_1)(y-y_2)$$
.
Solution:
Let (h,k) be the centre of the circle.
Now in the circle touching both the axes we have the equation as:
$$(x-h)^2+(y-h)^2=h^2$$
Where h=r=k so,
$$(x-2)^2+(y-2)^2=2^2$$
$$x^2+y^2-12=0$$Is the required equation
Solution:
We can assume the equation of circle as :
$$x^2+y^2+2gx+2fy+c=0$$
Since it passes through (1,0), (2,-2), (3,1) so it must satisfy the above quation.Hence substituting the equation in corresponding coordinates we get
$$2g+c=1$$
$$-4f+4g+c=8$$
$$2f+6g+c=-10$$
Solving this we get,
$$g=\frac{-5}{2}$$
$$f=\frac{1}{2}$$
$$c=4$$
Hence equation of the circle is
$$x^2+y^2-5x+y+4=0$$
Solution:
Let O(h,k) be centre of the circle.Since the circle touches x-axis so radius of circle(r)=k and also from given coordinate h=3
Now equation of circle is:
$$(x-3)^2+(y-k)^2=k^2...(i)$$
Since the circle (i) passes through the point (1,2), so
$$4+(2-k)^2=k^2$$
$$or, 4k=8$$
$$Therefore,k=2$$
Now required equation of circle is
$$(x-3)^2(y-2)^2=2^2$$
$$Therefore,x^2+y^2-6x-4y+9=0 $$
is the required expression
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