A circle is the locus of points which moves in a plane such that its distance from a fixed point is always constant. The fixed point is called the ‘centre’ while the fixed distance is called the ‘radius’. This article helps you to have a clear idea of the topics such as circle, equation of tangent, normal and chord of contact.
Equation of Circle
1. Standard Equation of Circle:
x2+y2=r2
centre (0, 0) and Radius (r)
2. Equation of circle in centre radius form:
(x−h)2+(y−k)2=r2
Centre (h, k), Radius = r
3. Equation of circle in General form:
x2+y2+2gx+2fy+c=0
Where (–g, –f ) centre
r2 = g2 + f2 – c .
Radius = g2+f2−c
4. Equation of circle with points P(x1, y1) and Q(x2, y2) as extremities of diameter is
(x – x1) (x – x2) + (y – y1) (y – y2) = 0
5. Equation of circle through three non-collinear points P(x1, y1), Q(x2, y2) and R(x3, y3) is
We rewrite the given equation as x2+y2−38x−310y+1=0⇒g=−34,f=−35,c=1.
Hence the centre is (34,35) and the radius is
916+925−1=932=342.
Illustration 2: Find the equation of the circle with centre (1, 2) and which passes through the point (4, 6).
Solution:
The radius of the circle is (4−1)2+(6−2)2=25=5.
Hence the equation of the circle is
(x−1)2+(y−2)2=25⇒x2+y2−2x−4y=20.
Illustration 3: Find the equation of the circle whose diameter is the line joining the points (-4, 3) and (12, -1). Find also the length of intercept made by it on the y-axis.
Solution:
The required equation of the circle is
(x+4)(x−12)+(y−3)(y+1)=0.
On the y-axis, x=0⇒−48+y2−2y−3=0⇒y2−2y−51=0⇒y=1±52
Hence the length of intercept on the y-axis =252=413.
Illustration 4: Find the equation of the circle passing through (1, 1), (2, -1) and (3, 2).
Solution:
Let the equation be x2+y2+2gx+2fy+c=0.
Substituting the coordinates of the three given points, we get
2g+2f+c=−2,4g−2f+c=−5,6g+4f+c=−13.
Solving the above three equations, we obtain:
f=−1/2;g=−5/2,c=4.
Hence the equation of the circle is
x2+y2−5x−y+4=0.
Illustration 5: Find the equation of the circle whose centre is (3, 4) and which touches the line 5x+12y=1.
Solution:
Let r be the radius of the circle. Then
r = distance of the centre i.e. point (3, 4) from the line 5x+12y=1=∣∣∣∣25+14415+48−1∣∣∣∣=1362.
Hence the required equation of the circle is (x−3)2+(y−4)2=(1362)2⇒x2+y2−6x−8y+169381=0.
Illustration 6: Find the greatest distance of the point P(10, 7) from the circle x2+y2−4x−2y−20=0.
Solution:
Since S1=102+72−4×10−2×7−20>0, P lies outside the circle.
Join P with the centre C(2, 1) of the given circle.
Suppose PC cuts the circle at A and B, where A is nearer to C.
Then, PB is the greatest distance of P from the circle.
We have, PC =(10−2)2+(7−1)2=10
and CB = radius =4+1+20=5
∴ PB = PC + CB =(10+5)=15
Illustration 7: The tangent to circle x2+y2=5 at (1, -2) also touches the circle x2+y2−8x+6y+20=0. Find the coordinates of the corresponding point of contact.
Illustration 8: Find the angle between the two tangents from the origin to the circle (x−7)2+(y+1)2=25.
Solution:
Any line through (0, 0) be y−mx=0 and it is a tangent to circle (x−7)2+(y+1)2=25.
If 1+m2∣−1−7m∣=5⇒m=43,−34
The product of both the slopes is -1.
Hence, the angle between the two tangents is π/2.
Illustration 9: Find the equation of normal to the circle 2x2+2y2−2x−5y+3=0 at (1, 1).
Solution:
The centre of the circle is (1/2, 5/4)
Normal to circle at point (1, 1) is line passing through the points (1, 1) and (1/2, 5/4) which is x+2y=3.
Illustration 10: A foot of the normal from the point (4, 3) to a circle is (2, 1) and a diameter of the circle has the equation 2x−y=2. Then find the equation of the circle.
Solution:
The line joining (4, 3) and (2, 1) is also along a diameter.
So, the centre is the point of intersection of the diameter 2x−y=2 and the line y−3=4−23−1(x−4) i.e., x−y−1=0.
Solving these, we get the centre as (1, 0).
Also, the radius = the distance between (1, 0) and (2, 1) =2
Hence, the equation of circle is (x−1)2+(y−0)2=2or x2+y2−2x−1=0.
Illustration 11: Find the length of the common chord of the circles x2+y2+2x+6y=0 and x2+y2−4x−2y−6=0
Solution:
Equation of common chord is S1−S2=0 or 6x+8y+6=0 or 3x+4y+3=0.
Centre of S1 is C1(-1, -3) and its radius is 1+9=10..
Distance of C1 from the common chord =9+16∣−3−12+3∣=512
∴ Length of common chord =2(10)2−(512)2=210−25144=225106=52106
Illustration 12: A pair of perpendicular straight lines pass through the origin and also through the point of intersection of the curve with The set containing the value of a is
(a) {-2, 2} (b) {-3, 3} (c) {-4, 4} (d) {-5, 5}
Solution:
To make the curve x2+y2=4 homogeneous with respect to line x+y=a, we should have