Lines CA and CB touch the circle only at points A and B respectively,

Question

https://gmatclub.com/forum/download/file.php?id=54972 Lines CA and CB touch the circle only at points A and B respectively, as shown in the figure above. If the centre O of the circle is at the point (0,0), the coordinates of point C are (0,2) and the radius of the circle is √2 units, what are the coordinates of point A? A. (\frac{1}{\sqrt{2}}, \ \frac{\sqrt{3}}{\sqrt{2}}) B. (\frac{\sqrt{3}}{2}, \ \frac{\sqrt{5}}{2}) C. (1, \ 1) D. (\frac{\sqrt{5}}{2}, \ \frac{\sqrt{3}}{2}) E. (\frac{\sqrt{3}}{\sqrt{2}}, \ \frac{1}{\sqrt{2}}) 1.png

gurmukh · Answer

In right angle triangle ACO, OC =2, OA = sqrt2, therefore AC = sqrt 2
Now ACBO is rhombus, cutting OC and AB let's say at D.
In triangle ACO
1/2×2×AD= 1/2 × sqrt2×sqrt2
AD =1
In right angle triangle ODC
OD =1
Therefore coordinate of C are (1,1)

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Lipun · Answer

CAO is right angled at A.
OC=2 and OA= \sqrt{2}  => AC= \sqrt{2} .
AC=OA => AOC=45 degree.
Since, OA= \sqrt{2} , the only possible co-ordinate is (1,1).

DrVanNostrand · Answer

"Since, OA= \sqrt{2} , the only possible co-ordinate is (1,1). "

I think this approach is wrong as many answers will produce the same result of 2^1/2.

Lipun · Answer

Hi  prashantsharan ,

Your understanding is correct. There can be many solutions in general when we find the sides of a right triangle with hypotenuse  \sqrt{2} .

However , if your read the statement AC=OA => AOC=45 degree then you will come to understand that both AOC = BAO = 45.
Let the intersection point of AB and OC be D. ADO=90 and AOC = BAO = 45. ----(1)
From eqn(1), AD=OD. Since, both the sides are equal and in 1st quadrant, thus there can be only one such solution which satisfies OA=  \sqrt{2} .

Note: It is important to know where you can use the equal value approach.

Let me know if it helps.

Regards
Lipun

Ishatyagi100 · Answer

Can someone explain why we can't use the formula of distance between two points to solve this? If we take option B and calculate distance from O to A that gives sqrt2, which is equal to the radius given.

Jswat · Answer

Bunuel , can you provide a solution to this question?

Flower_Child · Answer

The fact that the distance is coming as the radius only proves that the point lies on the circle . But that doesn't prove that that point is A.
There will be many similar points points such as ( \frac{\sqrt{3}}{2}  ,  \frac{\sqrt{5}}{2}  ), ( \frac{\sqrt{4}}{2}  ,  \frac{\sqrt{4}}{2}  ) , ( \frac{\sqrt{6}}{2}  ,  \frac{\sqrt{2}}{2}  )
But only 1 point will be A which needs to be solved and not guessed in this question.

MclLaurent · Answer

Still waiting the explanation for this question.

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malimanju · Answer

1.Tangent meets a circle at 90°.
 So OAC =90° and CA=sqrt2 
2.Same way..
 OBC =90° and CB=sqrt2
Hence it's a Square.

Now diagonols of Square bisect each other(say at D)
 so CD=DO=1 (by this answer can be marked)
 And in right angle triangle ADO
 AD=1
Hence (1,1)

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RTN · Answer

Line A and B are tangent to the circle- tangent are perpendicular to radius AO and BO. Angle OAC and angle OBC= 90--(1)
Distance at which two tangents from an external point is same -CB=CA-- (2)

In triangle OAC- using Pythagoras- 2^2= √2^2+CA^2
CA=CB= √2

All sides are equal - √2

In triangle OBC and OAC are isosceles triangles- Angle BOC=Angle BCO= 45 degree.
In triangle OAC, angle AOC=angle ACO=45 degree.

Angles BCA. and BOC are also 90 degree

The guide is a square.

Squares have equal and bisecting diagonals. CO=BA=2

Diagonal CO and BA bisect each other at lets say D. CO/2=DC=DA=1

Coordinates are (1,1)

Sa800 · Answer

can someone provide a satisfying solution? there still seems to be a lot of arguments in the comments.

Flower_Child · Answer

This question can be solved using Pythagoras Theorem and formula for distance between two coordinate points .

Consider the triangle formed by one of the tangent arm, radius and the centre line along y axis joining the point outside circle to the centre of circle.
Since line joining centre of circle and tangent is perpendicular to tangent , we have a right angled triangle .
We know two sides of this right angled triangle .
Using Pythagoras theorem, we can find the length of tangent :
Length of tangent =  \sqrt{2}

Now equation of circle with centre (0,0) is  x^2 + y^2 = radius^2 
So,   x^2 + y^2 = 2   -------(1)

Also, the formula for distance between two coordinate points (x,y) and (a,b) is given by :
 (x-a)^2 + (y-b)^2 = distance^2 
Let (x,y) be the points at which tangent touches the circle.
We already know length of tangent= distance between (x,y) and (0,2) =  \sqrt{2}

i.e ,  (x-0)^2 + (y-2)^2 =(\sqrt{2})^2

x^2 +(y-2)^2 = 2

Substituting value of  x^2  from (1)

2 - y^2 + y^2 + 4 - 4y = 2

y=1

Answer = C

Lines CA and CB touch the circle only at points A and B respectively,

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Lines CA and CB touch the circle only at points A and B respectively,

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