# I'm a JEE 2010 aspirant [ and a student of ASKIITIANS ] Pls help me to solve these.Also pls tell me as to whther these are JEE- level questions when compared to current trends : 1. The curve x 2 -y-√5 x+1=0 interesects the x-axis at A and B . A circle is drawn passing theough A and B , then the length of tangent drawn from the origin to this circle is : [ans : 1 unit ] 2. A circle C and the circle x 2 +y 2 =1 are orthogonal and have radical axis parallel to y-axis , then C can be [ans : x 2 +y 2 +x+1=0 ] 3. If a circle S(x,y)=0 touches at the point (2,3) of the line x+y=5 and S(1,2)=0,then radius of such a circle is : [WHY CAN'T WE USE THE FORMULA c 2 = a 2 (1+m 2 ) ? ] [ans : 1/2 units ]

**I'm a JEE 2010 aspirant [ and a student of ASKIITIANS ] Pls help me to solve these.Also pls tell me as to whther these are JEE- level questions when compared to current trends :**

**1. The curve x ^{2}-y-√5 x+1=0 interesects the x-axis at A and B . A circle is drawn passing theough A and B , then the length of tangent drawn from the origin to this circle is : **

**[ans : 1 unit ]**

**2. A circle C and the circle x ^{2}+y^{2}=1 are orthogonal and have radical axis parallel to y-axis , then C can be **

**[ans : x ^{2}+y^{2}+x+1=0 ]**

**3. If a circle S(x,y)=0 touches at the point (2,3) of the line x+y=5 and S(1,2)=0,then radius of such a circle is : **

**[WHY CAN'T WE USE THE FORMULA c ^{2} = a^{2}(1+m^{2}) ? ]**

**[ans : 1/2 units ]**

## 1 Answers

Hi,

1.) **The curve cuts the x- axis at two points, obtained by putting y=0. The points are- [(√5-1)/2,0] and [(√5+1)/2,0]. Now the circle is drawn passing through these two points. Note that origin always lies outside the circle (since both intersection points are positive). Hence we can make use of the tangent secant theorem( If on a circle, a tangent 'PT' is drawn from external point 'P' and a secant is drawn from the same point, P to intersect circle at 2 points A and B, then PT ^{2}=PA.PB). Applying the same in this case, the origin is external point which intersects the circle at 2 points (solved above) and also a tangent is drawn from that point.**

**Let O be the origin, A be [(√5-1)/2,0] , B be [(√5+1)/2,0] and T be the tangent point.**

**Now, OT ^{2}= OA.OB**

** =>OT ^{2}= (√5-1)/2 x (√5+1)/2**

** =>OT ^{2}=4/4=1**

** =>OT = 1 {since distance is always positive.)**

**You can also solve this question by taking the points to be diametrically opposite one, since the question only mentions "a circle", so its a safe and simplifying assumption. After that you can use the diametrical form of a circle, (x-x _{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) =0. You get the equation of circle. And finally apply the formula T=√(S_{1}). , putting (0,0) in place of x and y in the equation and finally taking the root. You get the final answer as 1.**

2. **The second and third questions have a fault probably, since the answers you wrote, are not satisfying the conditions of the Questions. The answer of the second question is an imaginary circle and so it can't intersect the real circle given in the question orthogonally. **

**For the third question, the expression S(a,b) =0 means that the circle passes through (a,b) and satisfies the circle's equation. Hence according to the question the circle passes through (2,3) and (1,2). The radius of the circle is given to be 1/2 units and hence diameter is 1 unit. But when you find the distance between the two given points {(2,3) and (1,2)}, it comes out to be √2, which is greater than the diameter of the circle and hence not possible, since the diameter is the longest chord.**

**The second question is time taking and is easily solved only when the options are known. As far as the third ques. is concerned, it can be solved by three equations, two obtained by putting the given points and the third by finding the radius by using the formula, c ^{2}=a^{2}(1+m^{2}). You know m, you know c and hence we get a^{2}= 25/2.**

**Please check the question again and do post the correct one again, if it is some other answer there. We are always ready to help.**

**Thanks.**