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Kinshook
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
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Question Stats:
52% (02:31) correct
48%
(02:25)
wrong
based on 140
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If line L y=3+x forms tangent to a circle with center at (1,-4), what is the equation of the other tangent to the same circle parallel to line L?
A. y= x -11
B. y+x =13
C. y = x -13
D. y+x = 11
E. y = x + 5
A. y= x -11
B. y+x =13
C. y = x -13
D. y+x = 11
E. y = x + 5
RajatVerma1392
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Given, Line y =x+3 => x -y +3 = 0 and Center (1,-4)
The radius will be equal to the perpendicular distance from the center to the tangent:
radius = \(\frac{|1*1 + (-1)(-4) +3|}{(\sqrt{(1^2 + 1^2)})}\)
radius = 8 /\(\sqrt{2}\)
Radius = 4*\(\sqrt{2}\)
Let the line which will be parallel to the tangent: x-y+k = 0
So, the perpendicular distance to the new line from the center will also be equal to the radius.
Thus,
4*\(\sqrt{2}\) =\(\frac{|1*1 + (-1)(-4) +k|}{(\sqrt{(1^2 + 1^2)})}\)
upon sloving this:
|k+5| = 8
=> If k+5 = 8 => k = 3
=> If -k-5 = 8 => k= -13
Thus looking at the answer only C is the correct one.
So, the answer is C.
Please hit kudos if you like the solution.
The radius will be equal to the perpendicular distance from the center to the tangent:
radius = \(\frac{|1*1 + (-1)(-4) +3|}{(\sqrt{(1^2 + 1^2)})}\)
radius = 8 /\(\sqrt{2}\)
Radius = 4*\(\sqrt{2}\)
Let the line which will be parallel to the tangent: x-y+k = 0
So, the perpendicular distance to the new line from the center will also be equal to the radius.
Thus,
4*\(\sqrt{2}\) =\(\frac{|1*1 + (-1)(-4) +k|}{(\sqrt{(1^2 + 1^2)})}\)
upon sloving this:
|k+5| = 8
=> If k+5 = 8 => k = 3
=> If -k-5 = 8 => k= -13
Thus looking at the answer only C is the correct one.
So, the answer is C.
Please hit kudos if you like the solution.
General Discussion
Archit3110
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Kinshook
given y=3+x
so y intercept is 3 and m = 1 since we need to find the line ll to line L so its slope also has to be same or say 1
out of given options we can negate option B & D also since the y intercept of line L is +ve the ll line must have the y intercept either <+3 or -ve so option E also negated
the distance from the point of intersection of the line on the circumference of circle to the center (1,-4) will give radius of the circle so in this case the radius would be
√(1-0)^2+(-4-3)^2 ; ~ 7
we can now test with y intercept values for option A & C ; we see that at y= x -11 ; y intercept is -11 we get the radius value same as for y=3+x so IMO A is correct
Kinshook
could you please check and confirm why C is correct answer not A?
Kinshook
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Archit3110
Archit, would you please explain your calculation marked in red above?
Archit3110
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Kinshook
It's the distance formula which I have used to find radius from center 1,-4 to y intercept 0,3, which is given in line L y=3+X and the point of intersection of this line would be on the circumference so using distance formula from center radiua can be found..
Posted from my mobile device
It's the distance formula which I have used to find radius from center 1,-4 to y intercept 0,3, which is given in line L y=3+X and the point of intersection of this line would be on the circumference so using distance formula from center radiua can be found..
Kinshook
Posted from my mobile device
Kinshook
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Archit3110
Archit, would you please explain your calculation marked in red above?[/quote]
Posted from my mobile device[/quote]
Y-intercept of tangent may not necessarily lie on point of tangency. They may be different points. If you want to calculate the radius, try to find perpendicular distance from center (1,-4) to line y=x+3 which\(x-y+3=0 is |1+4+3|/\sqrt{1+1} = 8/\sqrt{2} = 4\sqrt{2}\).
If you calculate the distance from center (1,-4) to line x-y-13=0, you will get \(|1+4-13|/\sqrt{2} = 8/\sqrt{2} =4\sqrt{2}\)
For A to be correct, x-y-11=0 distance from center (1,-4) =\(|1+4-11|/\sqrt{2}= 6/\sqrt{2}= 3\sqrt{2}\)
IMO C
Archit3110
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Y-intercept of tangent may not necessarily lie on point of tangency. They may be different points. If you want to calculate radium, then try to find perpendicular distance from center (1,-4) to line y=x+3 which\(x-y+3=0 is |1+4+3|/\sqrt{1+1} = 8/\sqrt{2} = 4\sqrt{2}\).
If you calculate the distance from center (1,-4) to line x-y-13=0, you will get \(|1+4-13|/\sqrt{2} = 8/\sqrt{2} =4\sqrt{2}\)
For A to be correct, x-y-11=0 distance from center (1,-4) =\(|1+4-11|/\sqrt{2}= 6/\sqrt{2}= 3\sqrt{2}\)
IMO C[/quote]
Kinshook ; okay thanks , could you please share what is the formula which you have used highlighted part to find perpendicular distance , honestly I have never come across such a formula or question like this earlier so its a first for me also could you also please share source of the question .
If you calculate the distance from center (1,-4) to line x-y-13=0, you will get \(|1+4-13|/\sqrt{2} = 8/\sqrt{2} =4\sqrt{2}\)
For A to be correct, x-y-11=0 distance from center (1,-4) =\(|1+4-11|/\sqrt{2}= 6/\sqrt{2}= 3\sqrt{2}\)
IMO C[/quote]
Kinshook ; okay thanks , could you please share what is the formula which you have used highlighted part to find perpendicular distance , honestly I have never come across such a formula or question like this earlier so its a first for me also could you also please share source of the question .
Kinshook
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The line joining center of circle to tangent will be perpendicular to the tangent and will pass through the center and will intersect tangent at point of tangency. Distance between center and point of tangency will provide the radius. The same line will intersect another tangent at a point for which distance between center of the circle and point of tangency will also be the radius. In other words center will be in the middle of 2 tangent lines.
Formula for perpendicular distance from (x,y) to line ax+by+c=0 is \(|ax+by+c|/\sqrt{(a^2+b^2)}\)
This question is created by me.
Formula for perpendicular distance from (x,y) to line ax+by+c=0 is \(|ax+by+c|/\sqrt{(a^2+b^2)}\)
This question is created by me.
The distance between the parallel line and the centre of the circle called the shortest or perpendicular distance is given by |ax+by+c|/sqrt(a^2+b^2). The tangent to the circle can be anywhere on the circumference of the circle but we will find the perpendicular distance from the line to the center of the circle. Let me first say that the distance between the parallel lines will be twice that distance since that other parallel line is, well parallel and also has a shortest distance equal to the perpendicular distance between it and the circle's centre. So now let us put in our values : L y=3+x ,.rewriting L in the form ax+by+c=0 gives us y-x-3=0.From this a =-1 b=1 and c is 3. We will put the point (1,-4) in the equation to give us -4-3-1=-8. Now |ax+by+c|/sqrt(a^2+b^2) gives us |-8|/sqrt(1^2+1^2) giving us 8/sqrt(2).
I said the distance between the parallel lines would be twice this so multiplying the distance by 2 gives us 8sqrt(2). The distance between two parallel lines is |b-c|/sqrt(m^2+1) where m is their common gradient and b and c are their y-intercepts. We know the y intercept for Line y=3+x is 3 and we know their common slope is 1 Plug it all in to get |3-c|=8 Now we get the y intercept for second parallel line to be -13 or 19. We can form the equation for line 2 knowing its slope and its intercept at the y-axis. This gives us y=x-13 or y=x+19. Only C has that.
I said the distance between the parallel lines would be twice this so multiplying the distance by 2 gives us 8sqrt(2). The distance between two parallel lines is |b-c|/sqrt(m^2+1) where m is their common gradient and b and c are their y-intercepts. We know the y intercept for Line y=3+x is 3 and we know their common slope is 1 Plug it all in to get |3-c|=8 Now we get the y intercept for second parallel line to be -13 or 19. We can form the equation for line 2 knowing its slope and its intercept at the y-axis. This gives us y=x-13 or y=x+19. Only C has that.
Kinshook
Equation of line: y=mx+b
Parallel lines have the same slope; so, m=1 for L and its parallel
Radius is the shortest distance between the center of the circle and line L
Shortest distance between a line and a point:
\(\frac{|ax+by+c|}{\sqrt{aˆ(2)+bˆ(2)}}\)
x and y are the coordinates of the circle (x=1,y=-4)
Line L y=3+x in the correct form becomes x-y+3=0 (a=1,b=1,c=3)
\(radius=\frac{|(1)(1)+(-1)(-4)+3|}{\sqrt{1ˆ2+(-1)ˆ2}}=|8|/\sqrt{2}=4*\sqrt{2}\)
This distance must be the same for the parallel line: y=x+b, x-y+b=0;
\(4*\sqrt{2}=\frac{|(1)(1)+(-1)(-4)+b|}{\sqrt{1ˆ(2)+(-1)ˆ(2)}}\)
\(4*\sqrt{2}*\sqrt{2}=|1+4+b|…8=|b+5|…b=[8-5=3,-8-5=-13]\)
The y-intercept of the parallel line must be a negative, so b=-13
Ans (C)
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