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shasadou
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Circle ABCD in the diagram above is defined by the equation x^2+y^2=25. Line segment EF is defined by the equation 3y=4x+25 and is tangent to circle ABCD at exactly one point. What is the point of tangency?
A. (–4, 3)
B. (–3, 4)
C. (–4, 7/2)
D. (–7/2, 3)
E. (–4, 4)
Attachment:
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ENGRTOMBA2018
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shasadou
Property of 2 mutually perpendicular lines with slopes m1 and m2 ---> m1*m2=-1
Also, the radius drawn from the center of a circle to the point of tangency, is perpendicular to the tangent at the point of tangency.
Thus, equation of line perpendicular to 3y=4x+25 (slope = 4/3)---> y = mx+c with m*(4/3)=-1---> m = -3/4 and it passes through (0,0) giving you c=0 .
Thus the equation of line perpendicular to the given line ---> y = -3/4 * x . Now solve this equation with 3y=4x+25 to get the point of tangency as (-4,3)
Thus, A is the correct answer.
HKD1710
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The point of tangency would satisfy both the equations.
Only (–4, 3) does that. Beware of choice B
Option A is the correct choice!
Only (–4, 3) does that. Beware of choice B
Option A is the correct choice!
General Discussion
We are given the equation of a line: 3y = 4x + 25 => y = (4/3)x + 25/3
Substituting the value of y in the equation of circle, we get:
x^2 + [ (4/3)x + 25/3 ]^2 = 25
=> x^2 + (16/9)x^2 + (200/9)x + 625/9 = 25
Multiplying by 9 on both sides,
9x^2 + 16x^2 + 200x + 625 = 225
25x^2 + 200x + 400 = 0
x^2 + 8x + 16 = 0
This gives us the value of x = -4
Substituting this value in the equation of line we get,
3y = 4*(-4) + 25 = -16 + 25 = -9
So, y = -3
And out point is (x,y) = (-4,-3)
Ans: A.
Substituting the value of y in the equation of circle, we get:
x^2 + [ (4/3)x + 25/3 ]^2 = 25
=> x^2 + (16/9)x^2 + (200/9)x + 625/9 = 25
Multiplying by 9 on both sides,
9x^2 + 16x^2 + 200x + 625 = 225
25x^2 + 200x + 400 = 0
x^2 + 8x + 16 = 0
This gives us the value of x = -4
Substituting this value in the equation of line we get,
3y = 4*(-4) + 25 = -16 + 25 = -9
So, y = -3
And out point is (x,y) = (-4,-3)
Ans: A.
somtsat99
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Point should satisfy both the equations i.e, for circle and the line. Plug in the values to get the answers. Its A.
Great explanations all there by intelligent folks. I just looked up the coordinates 
That also serves some hints
That also serves some hints shasadou
There's a couple of formulas we should have at our disposal in order to solve this question; remember that the slope of the tangent line is the negative reciprocal of the slope of the radius of the circle. Well, we know the center of this circle (0,0). Hence
(x-0)^2 + (y-0)^2 = 25
The two values of x and y must satisfy this equation and also the equation 3y= 4x + 25
Thus
"A"
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Circle ABCD in the diagram above is defined by the equation x^2+y^2=25. Line segment EF is defined by the equation 3y=4x+25 and is tangent to circle ABCD at exactly one point. What is the point of tangency?
A. (–4, 3)
B. (–3, 4)
C. (–4, 7/2)
D. (–7/2, 3)
E. (–4, 4)
A point if tangent would satisfy both the equations of line and circle
\(x^2+y^2=25\) and \(3y=4x+25\)
The only point from the options that satisfy the equation is A \((–4, 3)\)
Hence, Answer is A
as the point of tangency is a common point, it must satisfy both the equations.
only option A and B satisfy the circle equation n this means these two points lie on the circumference of the circle.
now we can get the right answer by putting the values of the x and y co-ordinates on the tangent equation.
only option A satisfies this equation.
only option A and B satisfy the circle equation n this means these two points lie on the circumference of the circle.
now we can get the right answer by putting the values of the x and y co-ordinates on the tangent equation.
only option A satisfies this equation.
The options should satisfy the Equation of Circle thus eliminating c and d options.
The option a and b should satisfy equation of line thus eliminating option b.
The option a and b should satisfy equation of line thus eliminating option b.
Posts: 2
You can use this shortcut, from the equation of the circle it is certain that the radius is 5. Now the distance between the origin and the tangent is 5. Options A and B only satisfies this condition so eliminate all other options. Now between A and B, Option A satisfies the tangent line equation which means, point of intersection of tangent and circle is option A.
SangamJena
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We can solve the question by using distance formula as well.
Distance from the tangent point to the centre of the circle has to be 5 =radius.
Hence root { (-4_0)sq + (3-0)sq} = 5
Satisfied hence ans is A.
Distance from the tangent point to the centre of the circle has to be 5 =radius.
Hence root { (-4_0)sq + (3-0)sq} = 5
Satisfied hence ans is A.
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