10 - Co ordinate Geometry Questions Answers

Tangents are drawn to the circle x^2 + y^2=12 at the points where it is met by the circle x^2 +y^2 - 5x + 3y - 2=0.Find the point of intersection of these tangents.

Asked By: SUSHOVAN HALDAR
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Joshi sir comment

sovle the two eqs for x and y,

method : 

x^2 + y^2=12 

on putting this in 2nd eq. we get -5x+3y = -10 so x = (3y+10)/5

so [(3y+10)/5]+ y2= 12

now get y then x, then get the tangents in the first circle and then intersection point of the tangents.

ƒ(x)=|x-2| and g=ƒoƒ so g'(x)=.......? ; 2<x<4

Asked By: BONEY HAVELIWALA
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Joshi sir comment

for 2<x<4,   f(x) = x-2

so g(x) = f(x-2) = x-2-2 = x-4

so g'(x) = 1 

ƒ(x)=3|2+x| so ƒ'(-3)=? ,if its -3 , how?

Asked By: BONEY HAVELIWALA
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Joshi sir comment

ƒ(x)=3|2+x|

for x>-2,  f(x) = 3(2+x)  so  f'(x) = 3 

for x<-2, f(x) = -3(2+x) so f'(x) = -3

A(3,4) and B is a variable point on the line |X| = 6. Also, AB ≤ 4. Then the number of positions of B with integral co-ordinates is:

(a)5  (b) 6 (c) 10 (d) 12

 

Asked By: GAUTHAM GANESH
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Joshi sir comment

by diagram (6,4),(6,3),(6,2) are points having AB<4 besides it two upside points are also there by symmetry not shown in diagram. These points are (6,5) and (6,6)

Total no. of points = 5

 

For which integers n ≥ 3 does there exist a regular n-gon in the plane such that
all of its vertices have integer coordinates?
 
Asked By: BUDDHI PRAKASH
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Joshi sir comment

for n = 4, we can form a square with integral coordinates of all the vertices.

 

Prove that the four projections of vertex A of the triangle ABC onto the exterior
and interior angle bisectors of  B and  C are collinear.
Asked By: BUDDHI PRAKASH
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Joshi sir comment

in the given diagram red lines are angle bisectors (interior and exterior)and X, Y, Z, W are the projections of A on these red lines

by angle bisector property all these 4 points will lie on the line BC (either externally or internally) so these points will be in a line

If the lines represented by 2x²-5xy+2y² =0 , be the two sides of a || gm and the line  5x+2y=1 be one of teh diagonals. then  the equation of the other diagonal is ?

Asked By: AMIT DAS
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Joshi sir comment

2x²-5xy+2y² =0

=> (x-2y)(2x-y) = 0 

these 2 lines pass through origin so one of the diagonal will definitely pass through origin. Given diagonal does not contain origin so second diagonal will pass through origin.

For finding any other point in the second diagonal first solve the two lines with given diagonal one by one. After calculating the two vertax use the formula for middle point of given diagonal. This point will be present in the second diagonal also. 

 

Tips about circle

Asked By: ADMINISTRATOR
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Solution by Joshi sir

 

Circle - Important Definitions

 
 
 
 
Director circle: the locus of the point of intersection of two perpendicular tangents to a given conic is known as its director circle.
Chord of contact: the chord joining the points of contact of the two tangents to a conic drawn from a given point, outside it, is called the chord of contact of tangents.

Pole and Polar: 
Polar of a point with respect to a circle: through a point P(x1,y1) (inside or outside a circle) there be drawn any straight line to meet the given circle a Q and R, the locus of the point of intersection of the tangents at Q and R is called the polar of point P and P is the called the pole of the polar.

Diameter – definition as a locus: the locus of the middle points of a system of parallel chords of a circle is called a diameter of the circle.

Common chord of two circles: The chord joining the points of intersection of two given circles is called their common chord.

Angle of intersection of two curves: If the two curves C1 and C2 intersect at a point P and PT1 and PT2 be the tangents to the two curves C1 and C2 respectively at P. Then the angle between the tangents at P is called the angle of intersection of the two curves at the point of intersection.

Orthogonal curves: Two curves are said to intersect orthogonally when the two tangents at the common point are at right angles.

Radical axis: the radical axis of two circles is the locus of a point which moves in such a way that the lengths of the tangents drawn from it to the two circles are equal.

Radical centre: The point of concurrence of the radical axes of three circles whose centres are non-collinear, taken in pairs, is called the radical centre of the circles.

Coaxial system of circles: A system of circles, every pair of which has the same radical axis is called a coaxial system of circles.
(i) Circle passes through origin 
So radius = a² = h²+k² 

(x-h)²+(y-k)² = h²+k² 

(ii)Circle touches the x axis 
C(h,k) centre, a = radius 
To satisfy a = k 
So equation is 
(x-h)²+(y-a)² = a² 

(iii)Circle touches the y axis 
C(h,k) centre, a = radius 
To satisfy a = h 
So equation is 
(x-a)²+(y-k)² = a² 

(iv) When the circle touches both axes 

then h = k = a 
(x-a)²+(y-a)² = a² 

(v) When the circle passes through the origin and centre is on x-axis. 
C(h,k) centre, a = radius 

As centre is on x axis y coordinate is zero. So k = 0. 
As circle is passing through origin a = h 
(x-a)²+ y² = a² 

(vi) When the circle passes through the origin and centre is on y-axis. 
C(h,k) centre, a = radius 

As centre is on y axis x coordinate is zero. So h = 0. 
As circle is passing through origin a = k 
x²+(y-a)² = a²
 
x²+y²+2gx+2fy+c = 0 

Centre of this circle = (-g,-f) 
Radius = √(g²+f²-c)
 
If (x1,y1) and (x2,y2) are coordinates of end points of the diameter 

then the equation of the circle is 
(x - x1)(x - x2)+(y - y1)(y- y2) = 0
 
Intercept of a circle is a line that is a chord which is part of x axis 

Intercepts for the circle x²+y²+2gx+2fy+c = 0 

length of intercept on x- axis = 2√(g²-c)(You get it by putting y = 0) 
length of intercept on y- axis = 2√(f²-c)(You get it by putting x = 0)
 
Is a point in the circle, on the circle or outside the circle 

If the point is P find distance between the centre of the circle C and point P. 
If the radius of the circle be R 

CP is greater than R implies point is outside the circle.

CP = R implies point is on the circle

CP is less than R implies point is inside.
 
Parametric equations of x² + y² = r² 

x = r cos θ, y = r sin θ 

Parametric equations of (x-a)² + (y-b)² = r² 

x = a + r cos θ, y = b + r sin θ
 
Equation of the circle: x² + y² = a² 

Equation of the line: y = mx+c 

A line does not intersect a circle if the length of the perpendicular to the line from the centre of the circle is greater than the radius of the circle. 
|c/√(1+m²)|>a 

A line intersects a circle if the length of the perpendicular to the line from the centre of the circle is less than the radius of the circle. 

|c/√(1+m²)|<a 
A line touches a circle if the length of the perpendicular to the line from the centre of the circle is equal to the radius of the circle. 

|c/√(1+m²)| = a
 
Equation of the circle: x² + y² = a² 

Equation of the line: y = mx+c 

A line intersects a circle if the length of the perpendicular to the line from the centre of the circle is less than the radius of the circle. 

If it intercepts, the length of the intercept is 

2√([[a²(1+m²)-c²]/(1+m²) ]

Tangent to a circle at a given point

 
Condition of tangency: 

The line y = mx+c is tangent to a circle x² + y² = a² if the length of the intercept is zero. 
That means 2√([[a²(1+m²)-c²]/(1+m²) ] = 0 
=> a²(1+m²)-c² = 0 
=> c = ±a√(1+m²) 


Slope form: 

The equation of a tangent of slope m to the circle x² + y² = a² is 
Y = mx±a√(1+m²) (Value of c from tangent condition). 
The coordinate of the point of contact are (±am/√(1+m²), ±a/√(1+m²) 


Point form: 

The equation of a tangent at the point (x1,y1) to the circle x² + y²+2gx+2fy+c = 0 is 

xx1 + yy1 +g(x+x1)+f(y+y1) +c = 0
 
If slope of the tangent is m, then the slope of the normal is –(1/m)
 
The length of a tangent from the point (x1,y1) to the circle x² + y²+2gx+2fy+c = 0 is equal to √( x1² + y1²+2gx1+2fy1+c)
 
Let the point be (x1,y1) and the circle be x² + y² = a² 

Two tangents can be drawn.

The tangent will be of the form y = mx+a√(1+m²) 
And the two values of m for the pair is to be found by solving the quadratic equation 
m²(x1²-a²) -2mx1y1 +(y1²-a²) = 0
 
The equation for pair of tangents from the point (x1,y1) to the circle x² + y²+2gx+2fy+c = 0 is given by 

(x² + y²+2gx+2fy+c) (x1² + y1²+2gx1+2fy1+c) = (xx1 + yy1 +g(x+x1)+f(y+y1) +c) ² 

Expressed as SS’ = T²
 
Director circle: the locus of the point of intersection of two perpendicular tangents to a given conic is known as its director circle.


Equation of director circle of the circle x² + y² = a² is x² + y² = 2a²
 
The equation of the chord of contact of tangents drawn from a point (x1,y1) outside the circle to the circle x² + y² = a² is xx1+yy1 = a².
 
Polar of a point with respect to a circle: If through a point P(x1,y1) (inside or outside a circle) there be drawn any straight line to meet the given circle a Q and R, the locus of the point of intersection (T) of the tangents at Q and R is called the polar of point P and P is the called the pole of the polar.

Polar is the locus of point and pole is a point with respect to which polar is determined.



Equation to the polar of the point (x1,y1) w.r.t. to the circle x² + y² = a² is 

xx1+yy1= a² 

The polar of the point (x1,y1) w.r.t. to the circle x² + y²+2gx+2fy+c = 0 is given by 
(xx1+ yy1 +g(x+x1)+f(y+y1) +c) = 0 
The equation is same as the equation for the tangent to the circle at a point (x1,y1) on the circle.
 
The equation of the chord of the circle x² + y²+2gx+2fy+c = 0 bisected at the point (x1,y1) is given by 

T = S’ 
(xx1 + yy1+g(x+x1)+f(y+y1) +c) = x1² + y1²+2gx1+2fy1+c
 
Equation of the diameter bisecting parallel chords y =mx+c ( c is a parameter i.e., varies to give various chords) of the circle x² + y² = a² is x+my = 0

 

 

 

The radius of the largest circle, which passes through the focus of the parabola y2=4(x+y) and contained in it is ???

Asked By: AMIT DAS
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Joshi sir comment

given equation can be written as (y-2)2= 4(x+1)  so coordinate of focus (a, 0) will be x+1 = 1 and y - 2 = 0 so x = 0 and y = 2

let the centre of largest circle inside parabola is (k, 2) so equation of that circle will be (x-k)2 + (y-2)2 = k2

on solving parabola and circle we get x+ (4-2k) x + 4 = 0 

for largest circle roots of this quadratic equation should be equal so b2= 4ac

solve and get k = 4 so radius will be 4

blushblushConsider a branch of the hyperbola  x²-2y²-2√2x-4√2y-6=0, with vertex at the point A. Let B be one of the end points of its latus rectum .If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is ??????smiley

Asked By: AMIT DAS
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Joshi sir comment

 

x²-2y²-2√2x-4√2y-6=0 

Arrange this equation in form of standard hyperbola as 

(x-√2)2/4 - (y+√2)2/2 = 1

so X = x-√2, Y = y+√2

vertex coordinate :  X = 0 and Y = 0 so x = √2, y = -√2

similarly focus : X = ae, Y = 0,     here a = 2, b = √2 and b2 = a2(1-e2)

and end point of latus rectum : X = ae, Y = b2/a

solve the area and get the answer

 

 

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