Circles Theory

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Circles theory for IIT-JEE.Great help for maths co-ordinate geometry.

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<ul><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 1</p><p>Mathematics / Circles</p><p>Circles</p><p>As we know very well from pure geometry, a circle is a geometrical figure described by a moving point in theEuclidean plane such that its distance from a fixed point is always constant. The fixed point is called the centre ofthe circle while the fixed distance is called the radius of the circle.</p><p>Moving over to a co-ordinate system, let us denote the centre C of the circle by (x0, y0) and the radius by r. Forany point P(x, y) lying on the circle, the length PC must be equal to r. Using the distance formula for PC, wetherefore obtain:</p><p>( ) ( )2 2 20 0 : Equation of thecirclex x y y r + =</p><p>This equation must be satisfied by every point P(x, y) lying on the circle; therefore, this is the equation that uniquelydescribes the given circle. We simply call it the equation of the circle, with centre (x0, y0) and radius r.</p><p>For example, consider the circle with its centre at (1, 1) and radius equal to unity:</p><p>Fig - 01</p><p>(1,1)</p><p>y</p><p>A circle with centre(1 1) and 1, r = </p><p>x</p><p>The equation of the this circle is</p><p>( ) ( )2 2 21 1 1x y + =2 2 2 2 1 0x y x y + + =</p><p>Find the equation of the circle passing through the points (0, 0) , (3, 0) and (1, 2).</p><p>Section - 1 EQUATIONS DESCRIBING CIRCLES</p><p>Example 01</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 2</p><p>Mathematics / Circles</p><p>Solution: To write the equation of the required circle, we must find its centre and radius.</p><p>Recall from pure geometry that a circle can always be drawn through three non-collinear points. Thiscan be done as follows: join the points to form a triangle. Draw the perpendicular bisectors of any ofthe two sides of this triangle. Their point of intersection gives us the centre C. The distance of C fromany of the vertices gives the radius r of the circles:</p><p>C is the centre of the circle passing</p><p>Fig - 02</p><p>C</p><p>P</p><p>Q R</p><p>through P, Q, R.</p><p>CP = CQ = CR = r.</p><p>We apply this result to the current example:</p><p>The equation of the perpendicular bisectorof OB is</p><p> 32</p><p>x =</p><p>The equation of the perpendicular bisector</p><p>y</p><p>xO (0,0) B (3,0)</p><p>C</p><p>A (1, 2)</p><p>Fig - 03</p><p>of OA is1 11 22</p><p>y</p><p>x</p><p>= </p><p>12 22</p><p>y x = </p><p>52 02</p><p>x y + =</p><p>The point C is the intersection of the two angle bisectors</p><p>3 1,2 2</p><p>C We can now easily evaluate the radius r as the length OC:</p><p>9 1 104 4 2</p><p>r OC= = + =</p><p>Finally, the equation of the required circle becomes:2 23 1 10</p><p>2 2 4x y + = </p><p>We will subsequently see another method to solve this type of questions. </p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 3</p><p>Mathematics / Circles</p><p>Find the equation of the circle which touches the co-ordinate axes and whose centre lies on the line 2 3x y =</p><p>Solution: A circle of radius r touching the co-ordinate axes can be in one of the four following configurations,with four corresponding equations mentioned alongside:</p><p>Fig - 04</p><p>( , )r r</p><p>y</p><p>x( ) + x r 2 ( ) = y r r2 2</p><p>( , )r r</p><p>y</p><p>x( ) + x + r 2 ( ) = y r r2 2</p><p>( )r r</p><p>y</p><p>x</p><p>( ) + x r 2 ( ) = y + r r2 2</p><p>( , )r r </p><p>y</p><p>x</p><p>( ) + x + r 2 ( ) = y + r r2 2</p><p>Note from these four possible cases that the centre of such a circle either lies on y = x or on y = x.</p><p>In the current example, the centre is also given to lie on x 2y = 3. Thus, there will be two circles,with the two centres being given by the point of intersection y = x and y = x with x 2y =3.</p><p>( )1and 2 3 3,3y x x y C= = Equation of the circle is ( ) ( )2 23 3 9x y+ + =</p><p>( )2and 2 3 1, 1y x x y C= = Equation of the circle is ( ) ( )2 21 1 1x y + + = </p><p>Find the equation of the circle with radius 5 and which touches another circle 2 2 2 4 20 0x y x y+ = externallyat the point (5, 5)</p><p>Solution: Let us first try to rearrange the equation of the given circle in the standard form from which well beable to deduce its centre and radius:</p><p>2 2 2 4 20 0x y x y+ =</p><p>( ) ( )2 21 2 25x y + =</p><p>Example 02</p><p>Example 03</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 4</p><p>Mathematics / Circles</p><p>Therefore, the centre of this circle is (1, 2) and its radius is 5.</p><p>We should now draw a geometrical figure which will certainly make things more clear:</p><p>Fig - 05</p><p>y</p><p>x</p><p>( )x , y0 0 </p><p>(5, 5)</p><p>(1, 2)</p><p>5</p><p>The required circle</p><p>Let the centre of the required circle be ( , ). Observe that is the mid-pt of (1,2) and ( , )</p><p>x yP</p><p>x y</p><p>0 0</p><p>0 0P</p><p>The given circle</p><p>As explained in the figure, we can now evaluate (x0, y0), the centre of the required circle:</p><p>00</p><p>1 5 92</p><p>x x+ = = 0 02 5 8</p><p>2y y+ = =</p><p>Thus, the required equation is</p><p>( ) ( )2 29 8 25x y + = </p><p>EQUATION OF A CIRCLE : GENERAL FORM</p><p>Expanding the standard form of the equation of the circle we derived in the last section, well obtain:</p><p>2 2 2 2 20 0 0 02 2 0x y x x y y x y r+ + + =</p><p>This suggests that the most general form of the equation of a circle can be written in terms of three variables; callthem g, f and c so that</p><p>2 2 20 0 0 02 2 ; 2 2 ;g x f y c x y r= = = + </p><p>2 2 2 2 20 0 0 0; ;x g y f r x y c g f c = = = + = + </p><p>Thus, the equation of the circle in terms of g, f and c becomes</p><p>2 2 2 2 0x y gx fy c+ + + + = : Equation of a circle; most general form</p><p>( ) 2 2Centre , ; Radiusg f g f c= = + It should be apparent to you how the standard form and the general form of the circles equation are interconvertible.Which form to use where is a matter of convenience and will depend on the situation.</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 5</p><p>Mathematics / Circles</p><p>As a first example, let us redo Example - 1, which involves finding the equation of the circle passing through thepoints (0, 0), (3, 0) and (1, 2).</p><p>Let the equation be 2 2 2 2 0x y gx fy c+ + + + = , where g, f and c are to be determined. This equation must besatisfied by the three points through which the circle passes, and hence well obtain three equations from whichg, f and c can be determined:</p><p>Substitute (0, 0) : c = 0</p><p>Substitute (3, 0) : 9 + 6g = 0</p><p>32</p><p>g = </p><p>Substitute (1, 2) : 1 + 4 3 + 4f = 0</p><p>12</p><p>f = </p><p>The required equation is hence:</p><p>2 2 3 0x y x y+ =</p><p>which is the same as what we obtained in Example - 1.</p><p>Let C be any circle with centre ( )0, 2 . Prove that at the most two rational points can lie on C.Solution: By a rational point, we mean a point which has both its co-ordinates rational.</p><p>Let the equation of C be 2 2 2 2 0x y gx fy c+ + + + =</p><p>We can arrive at the result easily be contradiction. Suppose that we have three rational points on thecircle with the co-ordinates (xi, yi) i = 1, 2, 3. These three points must satisfy the equation of thecircle. Thus we obtain a system of linear equations in g and f :</p><p>( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )</p><p>2 22 2 1 1 1 11 1 1 12 2 2 22 2 2 2 2 2 2 22 2</p><p>2 23 3 3 33 3 3 3</p><p>2 22 2 02 2 0 2 22 2 0 2 2</p><p>x g y f c x yx y gx fy cx y gx fy c x g y f c x yx y gx fy c x g y f c x y</p><p> + + = ++ + + + = + + + + = + + = + + + + + = + + = + </p><p>The coefficients in this system of linear equations are all rational by assumption. Thus, when we solve</p><p>this system, we must obtain g, f and c to be all rational. But since the centre is ( )0, 2 , we have2f = which gives us a contradiction.</p><p>This means that our assumption of taking three rational points on the circle is wrong At the mosttwo rational points can lie on this circle.</p><p>Example 04</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 6</p><p>Mathematics / Circles</p><p>Suppose we are given two curves C1 and C2 whose equation are as follows:</p><p>2 21 1 1 1 1 1 1: 2 2 2 0C a x h xy b y g x f y c+ + + + + =</p><p>2 22 2 2 2 2 2 2: 2 2 2 0C a x h xy b y g x f y c+ + + + + =</p><p>It is also given that these curves intersect in four concyclic points. Prove that</p><p>1 1 2 2</p><p>1 2</p><p>a b a bh h </p><p>=</p><p>Solution: From the discussions in the last chapter, we know that any curve C passing through the point(s) ofintersection of two given curves 1 20 and 0C C= = can be written as</p><p>1 2 0 whereC C C + = !</p><p>We can do the same in the current example to obtain the equation of the curve passing through the four(concyclic) points of intersection as:</p><p>( ) ( ) ( ) ( ) ( )2 21 2 1 2 1 2 1 2 1 2 1 22 2 2 0a a x h h xy b b y g g x f f y c c+ + + + + + + + + + + =From the general form of the equation of the circle, we know that this equation (above) will representthe equation of a circle only if:</p><p>Coeff. of x2 = Coeff. of y2 1 2 1 2a a b b + = + </p><p>1 1</p><p>2 2</p><p>b ab a</p><p> = </p><p>... (1)</p><p>Coeff. of xy = 0 1 2 0h h + =</p><p>1</p><p>2</p><p>hh</p><p> = ... (2)</p><p>From (1) and (2), we have</p><p>1 1 2 2</p><p>1 2</p><p>b a b ah h </p><p>= </p><p>Suppose that the equation of a circle is</p><p>2 2 2 2 0S x y gx fy c + + + + =</p><p>What condition must the co-ordinates of a point P(x1, y1) satisfy so that P may lie (i) inside the circle (ii) outside thecircle?</p><p>Example 05</p><p>Example 06</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 7</p><p>Mathematics / Circles</p><p>Solution: Let the centre of S be C and its radius be r.</p><p>The point P lies inside S if CP &lt; r and outside S is CP &gt; r.</p><p>From the equation of S, we know C to be (g, f) and r to be 2 2 .g f c+ Using these facts, wecan easily evaluate the required conditions:</p><p>P lies inside the circle: 2 2CP r</p><p>( ) ( )2 2 2 21 1x g y f g f c+ + + &gt; + 2 21 1 1 12 2 0x y gx fy c+ + + + &gt; ... (2)</p><p>We can write (1) and (2) concisely as</p><p>( )( )( )</p><p>1 1</p><p>1 1</p><p>1 1</p><p>lies inside the circle , 0</p><p>lies on the circle , 0</p><p>lies outside the circle , 0</p><p>P S x y</p><p>P S x y</p><p>P S x y</p><p>Find the equation of the circle circumscribing the triangle formed by the lines 6, 2 4 and 2 5x y x y x y+ = + = + = .</p><p>Solution: Let us first consider the general case wherein weve been given three lines L1, L2 and L3 and we needto find the circle circumscribing the triangle that these three lines form</p><p>Fig - 06</p><p>O</p><p>C</p><p>BA</p><p>L3L2</p><p>L1</p><p>Example 07</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 8</p><p>Mathematics / Circles</p><p>One way to do it would be as follows:</p><p>! Find the intersection points A, B and C of the three lines! Use these intersection points to write any two perpendicular bisectors! Find the intersection of these two perpendicular bisectors which gives us the centre O.! Finally, find the radius (which will equal OA, OB and OC)</p><p>This procedure will definitely become quite lengthy. We look instead for a a more elegant method.We first try to write the equation of an arbitrary second-degree curve S passing through the intersectionpoints of L1, L2 and L3. Think carefully and youll realise that such a curve can be written in terms oftwo arbitrary constants and as follows:</p><p>1 2 2 3 3 1 0S L L L L L L + + =That such a curve S will pass through all the three intersection points can be verified by observing thatthe substitution of the co-ordinates of any of the three points in the equation above will make bothsides identically 0.One we have such a curve, we can impose the necessary constraints to make it a circle.Coming back to the current example, the equation of an arbitrary curve passing through the intersectionpoints of the three lines can be written as:</p><p> ( )( ) ( )( ) ( )( )6 2 4 2 4 2 5 2 5 6 0S x y x y x y x y x y x y + + + + + + + + =To make S the equation of a circle, we simple impose the following constraints:</p><p>2 2 2 1 2 2= + + = + + 2Coeff. of Coeff. ofx y</p><p>1 = .. (1)</p><p>3 5 3 0 + + =Coeff. of = 0xy</p><p>65 = ... (2)</p><p>We substitute and back into S to obtain the required equation as:2 2 17 19 50 0S x y x y + + =</p><p>As another example of following such an approach, suppose that we are given four straight lines andare told that they intersect at four concyclic points, as shown below:</p><p>D</p><p>L2</p><p>Fig - 07</p><p>L4</p><p>L1</p><p>L3</p><p>C</p><p>B</p><p>A</p><p>A, B, C, D are four concyclic points</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 9</p><p>Mathematics / Circles</p><p>What approach will you follow if youre told to find the equation of the circle circumscribing thisquadrilateral? Obviously, one can always proceed by explicitly determining the centre and the radiusof the said circle, but as in the previous question, a much more elegant method exists.</p><p>Convince yourself that any second degree curve S passing through A, B, C, D can we written as</p><p>1 3 2 4 0S L L L L + =</p><p>Observe carefully that the substitution of the co-ordinates of any of the four points A, B, C, D willmake both sides identically 0, implying that these four points lie on S. We now simply impose the</p><p>necessary constraint ( )on to make S represent a circle, thus obtaining S ! </p><p>Consider a circle of radius r centred at the origin:</p><p>2 2 2x y r+ =</p><p>A line y = mx + c either just touches this circle or intersects it in two distinct points. What condition must m and csatisfy?</p><p>Solution: What we need to do algebraically is solve the simultaneous system of equations</p><p>2 2 2x y r+ = ... (1)</p><p>y mx c= + ... (2)</p><p>and find the condition on m and c for this system to have two distinct roots.</p><p>We substitute the value of y from (2) in (1):</p><p>( )22 2x mx c r+ + =</p><p>( )2 2 2 21 2 0m x mc x c r + + + = ... (3)Since the line intersects the circle (or touches it) the discriminant of (3) cannot be non-negative sinceat least one real value of x must exist. Thus:</p><p>( )( )2 2 2 2 24 4 1m c m c r + 2 2 2 2 0c r m r </p><p>( )2 2 21c r m +This is the condition that m and c must satisfy.Incidentally, we also obtained the condition for tangency:</p><p>( )2 2 21c r m= +21c r m = +</p><p>Example 08</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 10</p><p>Mathematics / Circles</p><p>Thus, 21y mx r m= + will always be tangents to the circle x2 + y2 = r2, whatever be the value ofm. A particular case that you should observe here is that when ,m the equation becomes</p><p>2 2lim</p><p>1 1my m x r</p><p>m m </p><p>= + + </p><p>x r = </p><p>which means that the two tangents are vertical and touch the circle at the two end-points of thehorizontal diameter. This is intuitively obvious. </p><p>A point P moves in the Euclidean plane in such a way that ,PA PB= where A and B are fixed points and 0 &gt; .Find the locus of P.</p><p>Solution: The easiest case is when 1; = then PA = PB and P will hence lie on the perpendicular bisector of AB.We consider the case when 1. Let A and B be assigned the co-ordinates (a, 0) and ( a, 0) (forconvenience). This can always be done by an appropriate choice of the co-ordinate axes.</p><p>Now, let P have the co-ordinates (x, y). We have,</p><p>2 2 2PA PB= </p><p>( ) ( ){ }2 22 2 2x a y x a y + = + +( ) ( ) ( ) ( )2 2 2 2 2 2 21 1 2 1 1 0x y ax x a + + + =</p><p>( )( )</p><p>22 2 2</p><p>2</p><p>12 0</p><p>1x y a x a</p><p>+ + + =</p><p> ... (1)</p><p>This is obviously the equation of a circle centred at ( )( )</p><p>2</p><p>2</p><p>1,0</p><p>1</p><p>a + . Note that this circle does not</p><p>pass through either A or B.</p><p>Let us consider an example of this. Let AB be 2 units, so that we can assign (1, 0) and (1, 0) as theco-ordinates A and B. Let P move in such a way that PA = 2PB, i.e, 2. = From (1), the locus of Pis the circle:</p><p>( )2 2 2 1 4 1 01 4</p><p>x y x+</p><p>+ + =</p><p>2 2 10 1 03</p><p>x y x + + + =</p><p>Example 09</p></li><li><p>LOCUSLOCUSLOCUSLOCUSLOCUS 11</p><p>Mathematics / Circles</p><p>The centre of this circle is 5 ,03</p><p> and its radius is ( )</p><p>225 40 1</p><p>3 3 + = </p><p>Fig - 08</p><p>y</p><p>A</p><p>P</p><p>(1, 0)B</p><p>(1, 0)5 ,03</p><p>For any point taken on the circumference of this circle, we will have 2</p><p>P</p><p> PA = PB</p><p>The circle that we obtained</p><p>A fixed line L1 intersects the co-ordinate axes at P(a, 0) and Q (0, b). A variable line L2 , perpendicular to L1,intersects the axes at R and S. Show that the locus of the points of intersection of PS and QR is a circle.</p><p>Solution: The equation of L1, using intercept form, can be written as</p><p>1x ya b+ =</p><p>bx ay ab + =</p><p>Since L2 is perpendicular to L1, its equation can be written as</p><p>2 0L ax by + =</p><p>where is a real parameter.</p><p>Using the equation of L2, we can...</p></li></ul>