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Line K is tangent to the circle with center (0, 0) at (-3, y) such tha

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Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:00
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Line K is tangent to the circle with center (0, 0) at (-3, y) such that its y-intercept is positive. If the radius of the circle is 5, what is the slope of Line K?

A. $$-\frac{4}{3}$$

B. $$-\frac{3}{4}$$

C. $$\frac{3}{4}$$

D. $$\frac{1}{2}$$

E. Cannot be determined

 This question was provided by Crack Verbal for the Game of Timers Competition

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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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19 Jul 2019, 01:58
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Bunuel wrote:
Line K is tangent to the circle with center (0, 0) at (-3, y) such that its y-intercept is positive. If the radius of the circle is 5, what is the slope of Line K?

A. $$-\frac{4}{3}$$

B. $$-\frac{3}{4}$$

C. $$\frac{3}{4}$$

D. $$\frac{1}{2}$$

E. Cannot be determined

 This question was provided by Crack Verbal for the Game of Timers Competition

OFFICIAL EXPLANATION: FROM CRACK VERBAL:

Since the circle is centered at (0, 0), has a radius of 5 and is tangent at (-3, y), using the distance formula we can find the value of y.

5^2 = (-3 – 0)^2 + (y – 0)^2 ---> 25 = 9 + y^2 --> y^2 = 16 --> y = 4

Now to find the slope of line K ideally we will require two points on the line, but we know that the radius and the tangent in a circle are always going to be perpendicular to each other. If two lines are perpendicular, the product of the slopes will always be equal to -1.

Here line K and the radius are perpendicular to each other and we have two points on the radius i.e. (0, 0) and (-3, 4). The slope of the radius will be -4/3. Since line K is perpendicular to the radius, the slope of line K will be ¾.
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:18
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Slope of two perpendicular lines is -1.

Distance between the points is 5 , hence solving for y

(-3) ^ 2 + y^2 = 5^2

y = 4

Hence the equation of line is

Since y=mx+c , using the points 0,0 and -3,4 we get m = -4/3. Hence slope of line will be 3/4. Option C
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Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:11
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Radius is given to be 5.

Tangent would meet the circle at a point such that it is perpendicular to the radius drawn from the centre to the point of intersection.

Given that the tangent meets the circle at (-3,y).

Using distance formula (0-(-3))^2 + (0-y)^2 = 5^2
y^2 = 16
y = +/- 4

Slope of radius line= (y2-y1)/(x2-x1) = (0+/-4)/(0-3)
=+/- 4/3

Slope of tangent*Sloe of radius line = -1
Slope of tangent = 3/4 or -3/4

So, it cannot be determined
Option E.
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:17
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(x−a)^2+(y−b)^2=r^2

(-3-0)^2+(y-0)^2=5^2
9+y^2=25
y^2=25-9=16
y=+4/-4
As Y-intercept in positive,y=4
slope=-4/3
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:20
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since centre of circle (0,0) radius=5
equation of circle $$(x-a)^2+(y-b)^2$$=$$r^2$$ (where a,b centre of circle and r radius)
$$x^2+y^2=25$$
line k is tangent at (-3,y) so this point lies on circle
now putting values of x and y in equation
9+$$y^2$$=25
$$y^2$$=16
y has two values +4 and -4
we will take y=4 ( since given that line k has positive y intersept)
now we can find the slope
points are (-3,4)
slope=tan θ (θ is angle)=$$\frac{y}{x}$$=$$\frac{4}{-3}$$
A is the answer
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Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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Updated on: 01 Jul 2019, 08:24
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Given - Point of tangent to the circle with center at O(0,0) is A(-3,y) and the Radius is 5.

So OA^2 = (0 - (-3))^2 + (0 - y)^2 = 5^2
Gives y=4 (Not -4 since y intercept is positive)

Let slope of OA be m1 = (4-0)/(-3-0) = -4/3

Slope of tangent is m2 = -1/m1 (Property of slopes of perpendicular lines) = 3/4

Answer C

Originally posted by taransaurav on 01 Jul 2019, 08:23.
Last edited by taransaurav on 01 Jul 2019, 08:24, edited 1 time in total.
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:23
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We need to establish 2 baselines here : If the line K is tangent to the circle,It is perpendicular to the radius at the point of tangency. If the y intercept of line K is positive it means Line K exists in Quadrant I and II and maybe Quadrant III. To find y we find the distance between the centre of the circle and the point (-3,y). Which gives us y^2=16. y could be 4/-4 but because line K is tangent to the circle in Quadrant I, y is 4. Now we need to find the slope of the radius which will be the negative reciprocal of line K. Finding the slope of the radius/line with points (0,0) and (-3,4) gives -4/3.The negative reciprocal of this will be 3/4.Thus C
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Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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Updated on: 01 Jul 2019, 08:24
Since the line is tangent to the circle with center at (0,0) the distance between the two points is equal to radius; Squareroot[(y-0)^2+(-3-0)^2]=5; squaring on both sides gives y^2+9=25; y^2=16; y=4. Slope of line -4/3 (using (y2-y1)/(x2-x1) ) IMO A
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Originally posted by Arvind42 on 01 Jul 2019, 08:24.
Last edited by Arvind42 on 01 Jul 2019, 08:24, edited 1 time in total.
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:24
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Line K is tangent to the circle with center (0, 0) at (-3, y) such that its y-intercept is positive. If the radius of the circle is 5, what is the slope of Line K?

A. $$-\frac{4}{3}$$

B. $$-\frac{3}{4}$$

C. $$\frac{3}{4}$$

D. $$\frac{1}{2}$$

E. Cannot be determined

Line K is tangent to the circle with center (0, 0) at (-3, y) such that its y-intercept is positive

Since y-intercept is positive--> line rises as it moves towards the right
(-3,y) lies on the circumference of the circle with radius 5
y= 4
x^2+y^2= r^2

Line K is perpendicular to line L(that joins point (-3,4) and center of circle(0,0)
Slope of line L = -4/3

Slope of line K = 3/4 (Product of slope of perpendicular lines is -1)

Answer C
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:25
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circle has radius of 5 unit.
Y Intercept is positive so we take that point is in the second quadrant. So this point where tangent touches point (-3,Y)
5^2 = (-3-0)^2+(Y-0)^2
Y = 4
so slope of line joining point (-3, 4) and (0,0) = difference of Y points/difference of X points = (Y1 - Y2)/(X1-X2) = (4 - 0)/(-3- 0)=-4/3
and line tangent to this point will be perpendicular to this line so slope of the line K will be inverse and opposite in sign to the tangent joinining line (-3,4 ) and (0,0) and that will be equal to = 3/4
so IMO ans is C
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Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:27
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Radius is 5 and touching point is (-3,y).
Here y can be +4 or -4.
Since y-intercept is positive, y will be 4.
So, line joining the touching point and center will have slop of -4/3. Tanget will be perpendicular to this line with slop 3/4.

Answer : C

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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:34
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Given that,

Radius of the circle = 5
Tangent K meets the circle with center (0, 0) at (-3, y)
According to Tangent Theorem, the radius from the center of the circle to the point of tangency is perpendicular to the tangent line.

Therefore, we can solve for y, using the Pythagoras theorem.
$$-3^{2}$$+ $$y^{2}$$ = $$5^{2}$$
$$y^{2}$$ = 16
y = 4
Note: Since the question specifives the y intercept is positive, we can ignore the negative value of y, i.e. y = -4

Now using the slope equation of a straight line, y = mx + c, where m is slope and c is the y-intercept, we can find the equation of the radius
The slope of the radius is $$\frac{-4}{3}$$

The product of the slopes of two perpendicular lines is -1
Therefore, (slope of radius) x (slope of line k) = -1

Slope of line k = $$\frac{3}{4}$$

The correct answer is option C. $$\frac{3}{4}$$
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:37
Radius is 5. Which means cordinate is (-3,4).

Using slope formula (y2 -y1)/(x2-x1)

(4-0)/(-3-0) = -4/3

Ans A

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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:38
Radius is 5. Which means cordinate is (-3,4).

Using slope formula (y2 -y1)/(x2-x1)

(4-0)/(-3-0) = -4/3

Ans A

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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:39
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The correct answer is C (3/4)

How?

Well, the tangent touches the circle at one point and is perpendicular to the radius. That point is given as (-3,y) and we also know that radius is 5 units.

Now for the radius to be 5 units . Let us consider the radius to be a line with 2 points A, B. A is the origin (0,0) and B is (-3,y). And distance between A and B is 5

Therefore,$$\sqrt{(x2-x1)^2 +(y2-y1)^2} =5$$
$$3^2 +y^2=5^2$$
$$y^2=16$$
$$y=4$$

Slope of line AB = $$\frac{(y2-y1)}{(x2-x1)}$$ = $$\frac{(4)}{(-3)}$$ = $$\frac{-4}{3}$$

Radius and tangent are perpendiculars

Product of slope of 2 perpendicular lines = m1*m2=-1

$$\frac{-4}{3} * (Slope of Tangent) =-1$$
Hence. The slope of Tangent =$$\frac{3}{4}$$
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:41
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Hi,

As given y>0, and tangent touches at (-3,y), and the radius = 5.
We can use the distance formula between points to find y.
$$\sqrt{{(-3-0)^2} + {(y-0)^2}}$$ = $$\sqrt{{5^2}}$$

This will give us y= +4 or -4 , since we know y> 0, so y= +4.

Now, the point is (-3,4). We can use the line equation between points:

(y-0) = $$\frac{(4-0)}{(-3-0)}$$ * (x-0)
y = (-4/3) x

As per the property of circle, the tangent should touch at 90-degree angle, so the slope formula is => m1 * m2 = -1
using the formula we get :
(-4/3) * m2 = -1
m2 = (3/4)

Ans = 3/4

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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:47
Since radius is given we can find Y
y= 4 or y =-4 but given Y intercept as positive so Y = 4

Slope can be found by (y-0)/(-3-0)= -4/3
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:50
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This is a good question. I spent a bit too much time (2 minute and 46 seconds) on it as I had hoped to solve the question directly.

The best approach is: (1) Figure out that the (-3, y) is actually (-3, 4)
(2) So, the slope for the line between (-3,4) and (0,0) is -4/3
(3) The answer is -1 / (-4/3) = 3/4.

So, the answer is C.
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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

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01 Jul 2019, 08:50
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Line K is tangent to the circle at the point (-3,y). Let’s call this point A.

The centre of the circle is C (0,0).

Let’s drop a perpendicular from A to the x-axis. The co-ordinate of the same is (-3,0). We’ll call this point B.

Joining the three points, we get a right angled triangle right angled at B. The sides are: AB = y, BC = 3, AC = 5 (radius)

Using Pythagoras theorem, we get y = 4. Hence, the point of tangency A is (-3,4).

The slope of the line AC is: (0-4)/(0+3) = -4/3

As line K is perpendicular to AC (property of tangency), the slope of K is the negative reciprocal, ie. 3/4.

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Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha   [#permalink] 01 Jul 2019, 08:50

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