Which of the following equations give a circle tangent to y-axis at (0

Question

Which of the following equations give a circle tangent to y-axis at (0, 2) and having the distance between x-intercepts equal to 3 units?

I.  x^2 + y^2 + 5x - 4y + 4 = 0

II.  x^2 + y^2 - 5x - 4y + 4 = 0

III.  x^2 + y^2 - 5x - 4y + 1 = 0

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

sumitkrocks · Answer

Rearranging terms :

1. (x+ \frac{5}{2} )^2 + (y-2)^2 =  \frac{(5}{2)} ^2

Center at (- \frac{5}{2}  , 2) and radius of  \frac{5}{2}  
X=0 ; (y-2)^2=0 i.e y=2 i.e tangent at (0,2) is valid
Also; y=0 to find x intercept x=-1 x=-4 Distance b/w intercepts is 3 units

equation 1 holds good

2. (x- \frac{5}{2} )^2 + (y-2)^2 =  \frac{(5}{2)} ^2
Center at ( \frac{5}{2}  , 2) and radius of  \frac{5}{2}  
X=0 ; (y-2)^2=0 i.e y=2 i.e tangent at (0,2) is valid
Also; y=0 to find x intercept x=1 x=4 Distance b/w intercepts is 3 units 
equation 2 holds good

3. (x- \frac{5}{2} )^2 + (y-2)^2 =  \frac{(37}{4)} 
Center at ( \frac{5}{2}  , 2) and radius of  \sqrt{\frac{37}{4}}  
Y axis is not the tangent as radius is > center coordinate distance, indicating circle cuts y axis at two points
Reject equation 3

IMO Answer is D

Bunuel  would you please suggest a quicker approach as it took me 3+ mins to get to the answer

Fdambro294 · Answer

The only way I can figure out how to do this question is to “complete the square” for each given quadratic, set up the standard equation for a circle, find the center and the X intercepts, and finally look to find the Relationship of the Circle to Point (0 , 2)

-I-
(X)^2 + 5X + (5/2)^2 + (Y)^2 - 4Y + 4 = (5/2)^2

(X + 5/2)^2 + (Y - 2)^2 = (5/2)^2

The radius is 2.5

The center is at: (-2.5 , 2) and

(1st) the X intercepts occur when Y = 0

(X + 5/2)^2 + (-2)^2 = (5/2)^2

(X)^2 + 5x + 25/4 + 4 = 25/4

(X)^2 + 5x + 4 = 0

(X + 4) (X + 1) = 0

X intercepts occur at:

X = -1 
And 
X = -4

Which is 3 units apart. This is satisfied.

Then, when you place the center of the circle at (-2.5 , 2) and see that the radius stretches exactly 2.5 units across horizontally to meet the Y Axis at (0 , 2), you will see that this in fact a Point of Tangency.

Thus, you can see that the Y Axis will be a Tangent Line to the circle at point (0 , 2)

Roman numeral I works.

After doing the similar procedure with Roman numerals II and III, you will find that II works and III does not.

D

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Lebor · Answer

Is there a Formular? How do we get to (5/2)^2?

JoaquinCerda · Answer

Very good explanation!
Thank you so much!

Fdambro294 · Answer

The general formula for any circle in the coordinate plane is the following:

(X - h)^2 + (Y - k)^2 = (radius)^2

Where (h , k) is the Coordinates for the Center of the Circle.

When you have a Quadratic expression of the form:

a(x)^2 + b(x)

You can “Complete the Square” by taking (1/2) of the coefficient (b), then squaring it, and then adding it to the expression.

Of course anything you add to the expression also has to be subtracted from the expression to keep the value the same.

For example:

Given:

(x)^2 - 10(x) + 4 = 0

You can complete the square by taking (1/2) of 10 and squaring it, and then adding that value to each side of the equation. In this example here:

(x)^2 - 10(x) + (5)^2 + 4 = (5)^2

(x - 5)^2 + 4 = 25

(x - 5)^2 = 21

You can do the same actions for each of the expressions in the Roman Numerals and set up the general equation for a circle.

For Roman numeral I, you already have (y)^2 - 4(y) + 4 ———-> (y - 2)^2

So for the remaining (x)^2 + 5x ——-> take (1/2) of 5 and square it and then subtract the same value so that the overall net effect on the expression is 0

(x)^2 + 5(x) + (2.5)^2 - (2.5)^2

(x + 2.5)^2 - 6.25

Put together the 2 parts and you have the general equation form for a circle in the coordinate plane.

(x + 2.5)^2 + (y - 2)^2 = (2.5)^2

You can google “ultimate quant guide gmat club” to see the  Bunuel  guide, which includes discussions on all these topics.

I just used the fractional form (5/2) instead of the decimal form of 2.5

Bambi2021 · Answer

I suspect the shortcut here was not intended. Or am I wrong? Simply plug in (0,2) in the expressions given. Only I and II will be equal to 0. There is no need to check for radius and x-intercepts to arrive at the answer.

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andreagonzalez2k · Answer

It can be false, you have to check if x-intercepts are 3 units apart.
For example

x^2+y^2−6x−4y+4=0

is tangent to y-axis at (0, 2)

but x-intercepts are not 3 units apart.

An easy way to calculate the distance between x-intercepts is sustituting y=0 and seeing that the diference of solutions in quadratic equation is

\frac{\sqrt{b^2-4ac}}{a}

In the attachment you can see  ec1  and  ec2  as the circles in the question and  ec3  as the circle I have posted.

Bambi2021 · Answer

Thats why I called it an "unintended shortcut". Test takers that can not solve it properly can simply plug in (0,2) in the expressions to arrive at the correct answer.

Dhananjayrai · Answer

Since y axis is tangent to circle/circles at (0,2), that means (0,2) lies on the circle/circles. Just put the values in the equations and you’ll get I and II.

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Adarsh_24 · Answer

I don’t think we need to make it into a circle equation format at all.
X intercept means y=0. Then solve quadratic equation for x. Check difference between points is 3.
Y intercept means x=0. Then solve quadratic equation for y. Since it’s a tangent. One point only which should be 2.

Aryan19852 · Answer

Is Co-ordinate imp for GMAT?

Which of the following equations give a circle tangent to y-axis at (0

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Which of the following equations give a circle tangent to y-axis at (0

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