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

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.

(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

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

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

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

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

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. \(\frac{3}{4}\)

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 =\(\frac{3}{4}\)

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

Please hit Kudos if you like the solution.

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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Since y-intercept is positive and the intersection point lies on the perimeter of the circle,
\((-3)^2 + y^2 = R^2\) - where R = 5

\((-3)^2 + y^2 = 25\)

Therefore, y = +4 as y-intercept is positive
.
Moreover, The tangent-Line1 and Line2 [formed by joining intersection-point A(-3, 4) and center(0,0)] are perpendicular to each other.
Thus, the product of their slopes is -1. - \([M1*M2 = -1]\)

Slope of Line2 = \(Y2 - Y1/ X2 - X1\)
\(4 - 0 / (-3) - 0\) = \(-4/3\)

Hence, slope of tangent-Line1 = \(3/4\)

Knowing that y intersect is (+ve) means that the slope is positive, so A & B are out.

The point of tangency between line K and the circle make a triangle with the origin and the x-intersect point (-3,0).
if the radius is 5, then the point of tangency is (-3,4) following a Pythagoras triple.

if the slope of the radius perpendicular to the tangent line k is 4/-3, then the slope of the tangent k is 3/4 (negative the inverse). C

Given:
Circle with center at (0, 0)
The radius of the circle: 5

From the above data,
The equation of the Circle formed is: \((x-0)^2 + (y-0)^2 = 5^2 => x^2 + y^2 = 25\)

Given:
Line K is a tangent to the above circle at (-3,y) such that its y-intercept is positive.

To find the value of y, we put the coordinate (-3,y) in the equation of our circle:
\(x^2 + y^2 = 25 => (-3)^2+y^2 = 25\) => y = +4 (or) -4
Since the y-intercept is positive, by looking at the figure formed we understand that y = 4
Thus, the point at which the Tangent intersects the circle is (-3,4). (Fig 2)

Finding the slope of the tangent:

The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.
Thus, if we draw a radius to like K they will be perpendicular to each other.

The product of the “slopes” of two perpendicular lines is -1
(Slope of the radius to the tangent)*(Slope of line K) = -1
=> \(\frac{(0-4)}{(0+3)}\)* (slope of line K) = -1
=>\(\frac{-4}{3}\)* (slope of line K) = -1
=> The slope of line K =\(\frac{3}{4}\)

Answer: C

Please find attached the video solution.

radius=5
5^2=3^2+y^2
y=+-4
Since Y intercept positive, y=4
From Image, tan Ѳ= tan(90-y)= coty=3/4
Ans C

Rule 1: Sketch the Circle with Center (0 , 0) out into the Second II Quadrant.

Since the Radius = 5, the Circle will Cross the X-Axis on the Left at (-5 , 0) and the Y-Axis on the Top at (0 , +5)


Rule 2: The radius drawn from the Center of the Circle to the Point of Tangency will be PERPENDICULAR with the Tangent Line


1st) Drawing a Radius from the Origin to the Point of Tangency where Line K touches the Circle


This must occur somewhere on the Arc with a radius of 5 that is drawn from Point (-5 , 0) to Point (0 , +5)


2nd) Draw a Vertical Line from the Point of Tangency at (-3 , y) down to the X-Axis such that it is Perpendicular to the X-Axis.

This Vertical Line and the Radius drawn to the Point of Tangency create a 90 Degree Right Triangle with the X-Axis from (-3 , 0) to the Origin


Hypotenuse = 5 = radius

Leg = Portion of X-Axis = 3 Units

Other Leg = Vertical Distance on the Y-Axis from X = -3 to the Point of Tangency ---

----- must = 4, since this is a Right Triangle Pythagorean Triplet of 3-4-5



3rd) Now that we know the Point of Tangency occurs at Point (-3 , y) = (-3 , 4) we must use 1 more Rule

RULE: Lines that are Perpendicular in the Coordinate Plane will have (-)Negative Reciprocal Slopes


Tangent Line K creates a 90 Degree Angle with the Radius drawn from Origin (0 , 0) to the Point of Tangency which we just determined is (-3 , 4)


Slope of this Radius = (Rise) / (Run) = (4 - 0)/ (-3 - 0) = - 4/3

the Negative Reciprocal of: - 4/3 -------- is + 3/4




Therefore, in order for Line K to be Tangent to the Circle at Point (-3 , 4), the Slope of Line K must = + 3/4

Answer -C-

Wonderful explanation, as always.

Please share if there is any alternative approach to this question. I was thinking about determining the coordinates of intercept of line with the Y and X axis, and from those points make exact same calculation (Y2-Y1)/(X2-X1). I understand that angles of 3x4x5 triangle are not very useful. Thanks in advance.