12 Days of Christmas GMAT Competition - Day 6: In the following figure

Let's consider the radius of the circle is r, Then the coordinates of the circle's center is(r,r).
The equation of the circle is \((x-r)^2+(y-r)^2 = r^2\)


Using Statement 1:
(1) The co-ordinate of A is (4, 8).

\((4-r)^2 + (8-r)^2 = r^2;\)
\(r^2-8r-16r+16+64 = 0;\)

So \(r^2 - 24r +80 = 0\)

r can be 4 or 20;

We need statement 2 to get one specific radius

Statement 2:
(2) The radius is greater than 15.

r is 20

IMO C

We have to find the radius of the Circle

Let the circle with Centre C
Let the circle touches Y axis at Y and X Axis at X
then X*Y=C^2 (Tangent Rule)
Case-1
(1) The co-ordinate of A is (4, 8) - Cannot Determine the Radius
Case -2
(2) The radius is greater than 15- Even if the radius is greater than 15 we cannot find the radius

Option E

Q. (x-t)2 + (y-t)2 = t2
1. (4-t)2 + (8-t)2 = t2 => gives us 2 values of t=4,20
2. t> 15

combining both t= 20

In the following figure both the co-ordinate axes are tangent to the drawn circle. If A is a point on the circumference of the circle, then what is the radius of the cube?

(1) The co-ordinate of A is (4, 8).
(2) The radius is greater than 15.


(1)
Equation of circle:
(4-r)^2 + (8-r)^2 = r^2
80+r^2-24r = 0

r = 4 or 20

Radius of the circle is 20. Radius of the smaller circle going through (4,8) is 4.

Sufficient

(2)
The radius is greater than 15.

This is not sufficient

A is the answer

Equation of a circle of radius r centred at (h,k) is given by:

\((x−h)^2+(y−k)^2=r^2 \)

Since the circle touches the axes, both the axes are tangents to the circle. From (h,k) , if we join the points of intersection of the circle with axes, these lines (also radii) will be perpendicular to the tangent. (Radius of a circle is perpendicular to the tangent at the point of contact).

Hence, the 2 lines will be parallel to the x and y axes. The line parallel to the y−axis will intersect x-axis at (h,0) and line parallel to x-axis will intersect y-axis at (0,k) .

Also from the diagram, h=k=r . Hence, the equation of required circle becomes

\((x−r)^2+(y−r)^2=r^2\)

Statement 1
Since this circle passes through (4,8)

\((4−r)^2+(8−r)^2=r^2\)

On solving we get \(r^2-24r+80\)=0
r=20 and r=4.

From Figure r>4, so r=20 only possible value.

From statement 2
r>15. Many possibilities exist.

So, r=20. IMO Choice A.

Equation of circle is \((x− a )^2+(y−b)^2=r^2\)

Both axis are tangent to the circle then we have \(a = b = r \)

our equation ⟹ \((x− r )^2+(y−r)^2=r^2\)

Statement 1: Now Point A (4,8) lies on the circle, therefore \((4− r )^2+(8−r)^2=r^2\)

Simplifying we get the equation \(r^2 - 24r - 80 = 0\) with the roots as r = 4 or 20

We know that Point A (4,8) lies on the circumference of the circle, therefore, the radius must be greater than 4. So the radius is circle is 20

Statement 1: Sufficient

Statement 2: Not Sufficient

Ans: A

case 1 :- The co-ordinate of A is (4, 8) --> not sufficient
case 2 :- The radius is greater than 15. --> not sufficient

combine 1&2 still not sufficient
ans E

The task at hand is solved pretty easily using the Pythagorean theorem and using the data provided under Condition 1.

I attach the image of the geometrical structure that's helpful - but basically, we should draw three radiuses from the centre: two to the tangent points of the circle and one leading straight to A.
As it's quite easy to notice from the drawing, if we know the coordinates of A, we also know that BC equals to radius minus A's height (so, BK - 8, or R - 8).
Same applies to the horizontal radius: CH is of course parallel and equal to the radius, so AC = HC - A's X-coordinate, or AC = HC - 4 = R - 4.

Now let's look at the triangle ABC:
\(AB^2 = AC^2 + BC^2\\
R^2 = (R-4)^2 + (R-8)^2\\
R^2 = R^2 + 16 -8R + R^2 + 64 - 16R\\
R^2 = 2R^2 - 24R + 80,\)
or we get a simple quadratic equation:
\(R^2 - 24R + 80 = 0\)

It's easy to distinguish, using the Vieta's formulas, that the roots are 4 and 20. So, we know that there are two options for the radius.

And now it looks tempting to use the Condition 2, which tells us to choose the option above 15, which is 20.
However, this condition is excessive, because if the radius equals to 4, than the point A would be placed in a completely different part of the circle - to be precise, on the top on the diameter drawn from point K in my scheme.
This is not possible for two reasons:

(1) we were given the drawing where the placement of A is quite obvious; and
(2) we actually obtained the equation using A the way it's drawn - so if it's in any other place, the whole geometrical solution wouldn't have made any sense.

Therefore, we don't need any help to understand that the only possible root for the radius is 20. So, Condition 1 is enough - answer A.

In the given figure both the co-ordinate axes are tangent to the drawn circle. If A is a point on the circumference of the circle as shown, then what is the radius of the circle?

(1) The co-ordinate of A is (4, 8).

Clearly insufficient as 2 possibilities exists as shown below -



(2) The radius is greater than 15.

We cannot calculate the radius of the circle with this information. Clearly insufficient.

Combining statements 1 and 2:

There is one unique circle that passes through point A with radius more than 15 units. SUFFICIENT!!

Answer: Option C

let radius be r, and let the point on which circle touches x axis be (r,0), and point on which circle touches y axis be (0,r)

a) by using distance formula we have
(4-r)^2+(8-r)^2= r^2
r^2-24r+80=0
r can be both 4, and 20

Insufficient

b) Not sufficient we have no info about r

combining 1, and 2 we know r will be 20

Correct Answer C

OA) A

In the given figure both the co-ordinate axes are tangent to the drawn circle. If A is a point on the circumference of the circle as shown, then what is the radius of the circle?

(1) The co-ordinate of A is (4, 8).
(2) The radius is greater than 15.

Attached the figure for the reference

statement I)

based on figure, distance between point A and the center of the circle will = \(\sqrt{(r-4)^2+(r-8)^2}\)

and this distance is also equal to the radius of the circle r

so \(\sqrt{(r-4)^2+(r-8)^2}\) = r

(r-4)^2+(r-8)^2 = r^2

solving this equation we can find the value of r = 20 or 4

but it is already given that point A is on the circumference of the circle its x coordinate is 4, so r cannot be 4

so based on that value of r = 20

statement I is sufficient

statement II)

r>15 based on this information nothing can be concluded for the radius of the circle

statement II is insufficient

Let the radius of the circle be 'r'.
The center of the circle would be (r,r) as the axes are the tangents

(1) The co-ordinate of A is (4, 8).

Now considering the point on the circumference, we have that the radius of the circle would also be the distance between the center of the circle to the point A(4, 8)

Now from the above we get
(r-4)^2 + (r-8)^2 = r^2
=> r^2 + 16 - 8r + r^2 + 64 - 16r = r^2
=> r^2 -24r + 80 = 0
=> (r-20)(r-4) = 0
=> r = 20 or r = 4
Hence we have two values of r, hence insufficient.

(2) The radius is greater than 15.
Clearly insufficient as infinite circles are possible

Now Combining (1) and (2) we get
r = 20
Unique value hence sufficient

IMHO Option C

In the given figure both the co-ordinate axes are tangent to the drawn circle. If A is a point on the circumference of the circle as shown, then what is the radius of the circle?

(1) The co-ordinate of A is (4, 8). - No mention of radius or coordinates of second point on the tangent
Let a be distance from 4 to tangent (that is the radius)
we have a+4
Insufficient

(2) The radius is greater than 15.
No mention of cordinates of the point of contact

Insufficient

Combining both statements we have a>11

Insufficient

Ans : E

This solution is incorrect. Statement 1 is not sufficient, as shown in the picture I've attached.