April 28, 2023 Schoolix
Conic Sections - I
# Equation of Circle
1) The equation of a circle centered at origin and with radius equal to a is
x ^ 2 + y ^ 2 = a ^ 2
2) The equation of a circle centred at (hk) and radius a is given by
(x - h) ^ 2 + (y - k) ^ 2 = a ^ 2
3) The general equation of a circle is
x ^ 2 + y ^ 2 + 2gx + 2fy + c = 0 where g ^ 2 + f ^ 2 - c > 0 , g, f, c ∈ R
4) The general equation of second degree
ax 2hxy by 2gx 2fy c = 0 in x and y
will represent a circle ta =b*0.h= h = 0 and a ^ 2 + f ^ 2 - ac > 0 a. bcfg .h in R
5) The equation of the circle on the line segment joining (x_{1}, y_{1}) and (x_{2}, y_{2}) as diameter is (x - x_{1})(x - x_{2}) + (y - y_{1})(y - y_{2}) = 0 M
6) The equation of the circle, in which the chord joining the points (x, y_{1}) and (x_{2}, y_{2}) subtends an angle 0 at the circumference of the circle is
(x - x_{o})(x - x_{o}) + (y - y_{o})(y - y_{o}) ± cot 0 * \{(x - x_{i})(y - y_{2}) - (x - x_{2})(y - y_{i})\} = 0
7) The equations of the circle with centre at (h, k) and touching x-axis and y-axis are
x ^ 2 + y ^ 2 - 2hx - 2ky + h ^ 2 = 0 and x ^ 2 * y ^ 2 - 2mx - 2ky + k ^ 2 = 0 respectively.
8) The equation of the circle touching both axes and radius 'a' is
(x + y ^ 2 ± 2) * ax + 2ay + a ^ 2 = 0
9) The equation of a circle passing through the point of intersection of a given circle S = 0 and a given line L = 0isS + lambda*L = 0 , where 2 is a constant to be determined by the given condition.
10) The equation of a circle passing through the point of intersection of two given circles S₁ = 0 and S_{2} = C is given by S_{1} + lambda*S_{2} = 0 , where lambda ne-1 and leading coefficients of x² and y in S, and S_{2} are unity.
# Intersection of a Straight line and a Circle
(1) A straight line L = 0 will intersect or touch or neither intersect nor touch the circle s = 0 according as the length of perpendicular drawn from the centre fS = 0I less than or equal or greater than the radius of the circle.
2) The lengths of the intersect made by the circle x2 + y2 + 2gx + 2fy + c = 0 on x-axis and y-axis are PQ = 2sqrt(g ^ 2 - c) and RS = 2sqrt(f ^ 2 - c) respectively
3) Let C be the centre a' be the radius of the circle and P be any pointThen P lies outside, inside or on the circle fCP > or < or = a .
# Equations of Tangent and Normal
1) The equation of the tangent and the normal to the circle x² y ^ 2 = a ^ 2 at (x_{1}, y_{1}) ^ 15 alpha_{1} + y*y_{1} =a^ angle and x/x_{1} - y/y_{1} = 0 respectively.
2) The equation of the tangent and the normal to the circle x2 + y ^ 2 + 2gx + 2fy + c = 0i (x_{1}, y_{1}) is xx, + y*y_{1} + g(x + x_{1}) + f(y + y_{1}) + c = 0 and (x - x_{1})/(x_{1} + g) = (y - y_{1})/(y_{1} + f) respectively.
3) The straight line y = mx + cu be tangent to the circle x ^ 2 + y ^ 2 = a ^ 2 if * c ^ 2 =a^ 2 ( 1 + m²). Then the equation of the tangent to the circle x ^ 2 + y ^ 2 = a ^ 2 in slope form may be written as
y = mx plus/minus a * sqrt(1 + m ^ 2) and the points of contact are
am
√1
# Image of a circle with respect to a line mirror
Let the equation of the circle be S = (x - h)2 + (y - k)2 - a2 = 0 and the line mirror be Ix + my + n = 0
Radius of the image circle will be same as the radius of S, but centre will be changed.
1) The equation of the image of the circle S = 0 with respect to x-axis as the line mirror is (x - h)² + (y + k)² = a²
2) The equation of the image of the circle S = 0 with respect to y-axis as the line mirror is (x + h)²+(y-k)² = a2
3) The equation of the image of the circle S = 0 with respect to the line Ix + my + n = 0 is (x - h')²+(y-K)² = a2
Where (h', K) is the image of (h, k) with respect to the line mirror Ix + my + n = 0
# Special Points
1) The area of equilateral triangle inscribed in x² + y2 + 2gx + 2fy + c = 0 is
3√3 4 (g² + f² - c)
2)Equation of circumcircle of triangle formed by the lines L,₁ = 0, L₂ = 0, L3 = 0 is 442 + 2ghz +µhigh = 0, where λ and μ are to be find out by using the concept that coefficient of x2 = coefficient of y2 and coefficient of xy = 0.
3)
In the case the equation of the circle passing through the points A, B, C, D is 44 +244₁ =0 where can be find out by using standard conditions such as coefficient of x² = coefficient of y2 and coefficient of xy = 0.
4)
Here the relation PA. PB = PC. PD = PT^2 always holds.
5)
If C is the Centre of circle, AB is the chord of contact of tangents from P to AB, r is the radius of circles PA = PB = l
Then, AB= 2rl/√r2 + l2
Area of triangle PAB= rl^3/r2 + l2
