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8.9D Perpendicular Lines and the equation of a tangent: Fluency and Exam Drill

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 60 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A circle has equation x² + y² = 25. Find the gradient of the tangent to the circle at the point (3, 4).

(Total for Question 1 is 1 mark)

For the circle x² + y² = 25, write down the equation of the tangent at the point (3, 4).

(Total for Question 2 is 2 marks)

Find the equation of the tangent to the circle x² + y² = 625 at the point (7, 24).

(Total for Question 3 is 2 marks)

The circle has equation (x - 1)² + (y + 2)² = 25 and P is the point (5, 1) on the circle.
Find the equation of the line through P that is perpendicular to the tangent at P.

(Total for Question 4 is 2 marks)

The circle x² + y² = 50 meets the vertical line x = 5 at two points. Take the point with positive y-coordinate.
Work out the equation of the tangent to the circle at this point.

(Total for Question 5 is 3 marks)

Show that for a circle x² + y² = a² the equation of the tangent at the point (x₀, y₀) on the circle is x x₀ + y y₀ = a².

(Total for Question 6 is 3 marks)

The circle x² + y² = 100 has a point A with x-coordinate 8 and y positive.
(a) Find the equation of the tangent at A.
(b) Find the coordinates of the point where this tangent meets the line y = 2x - 5.

(a)Find the equation of the tangent at A.(2)

(b)Find the coordinates of the intersection point of the tangent and the line y = 2x - 5.(2)

(Total for Question 7 is 4 marks)

Circle C has equation (x - 1)² + (y - 2)² = 25 and point P is (4, 6) on C.
(a) Find the equation of the tangent at P.
(b) Find the equation of the line through P that is perpendicular to this tangent.

(a)Find the equation of the tangent at P.(2)

(b)Find the equation of the line through P perpendicular to the tangent.(2)

(Total for Question 8 is 4 marks)

For the circle x² + y² = 25 find the point or points where the tangent has gradient 2.

(Total for Question 9 is 3 marks)

The circle x² + y² = 25 and the point P(4, 3) on it. The tangent at P meets the x-axis at A and the y-axis at B.
Work out the area of triangle OAB, where O is the origin.

(Total for Question 10 is 4 marks)

A circle has equation x² + y² = 50 and a point T lies on the circle. Show that the tangents from the external point (10, 0) touch the circle at points with x-coordinate 5, and find the coordinates of these points of contact.

(Total for Question 11 is 4 marks)

The circle x² + y² = 169 is given and the line L has equation 3x - 4y + 10 = 0.
(a) Find the points on the circle where the tangent is perpendicular to L.
(b) Find the equations of the tangents at those points.

(a)Find the points on the circle where the tangent is perpendicular to the line L.(4)

(b)Find the equations of the tangents at those points.(4)

(Total for Question 12 is 8 marks)

Mark scheme · 8.9D Perpendicular Lines and the equation of a tangent: Fluency and Exam Drill

Question 1

B1 gradient = -3/4 cao
Answer: -3/4 (or -0.75)

Question 2

M1 use tangent gradient -3/4 and substitute point to find intercept oe
A1 y = -3/4 x + 25/4 cao
Answer: y = -3/4 x + 25/4

Question 3

M1 use xx0 + yy0 = r² or find slope -7/24 then form equation oe
A1 7x + 24y = 625 cao
Answer: 7x + 24y = 625

Question 4

M1 find gradient of radius = (1 - (-2)) / (5 - 1) = 3/4 or identify perpendicular slope oe
A1 equation y - 1 = 3/4 (x - 5) or y = 3/4 x - 11/4 cao
Answer: y = 3/4 x - 11/4

Question 5

M1 find y: y² = 50 - 25 so y = 5 (positive) oe
M1 find tangent gradient -x/y = -5/5 = -1 oe
A1 y = -x + 10 cao
Answer: y = -x + 10

Question 6

dM1 use the radius vector (x0, y0) is perpendicular to tangent so (x0)(x - x0) + (y0)(y - y0) = 0 oe
M1 expand or rearrange to get x x0 + y y0 = x0² + y0² oe
A1 substitute x0² + y0² = a² to give x x0 + y y0 = a² cso
Answer: x x₀ + y y₀ = a²

Question 7

(a) M1 find y: y² = 100 - 64 so y = 6 and slope -x/y = -8/6 = -4/3 oe
(a) A1 y = -4/3 x + 50/3 cao
(a) Answer: y = -4/3 x + 50/3
(b) M1 solve -4/3 x + 50/3 = 2x - 5 oe
(b) A1 intersection (13/2, 8) cao
(b) Answer: (13/2, 8)

Question 8

(a) M1 radius slope = (6 - 2)/(4 - 1) = 4/3 so tangent slope = -3/4 oe
(a) A1 y = -3/4 x + 9 cao
(a) Answer: y = -3/4 x + 9
(b) M1 perpendicular slope = 4/3 and form y - 6 = 4/3 (x - 4) oe
(b) A1 y = 4/3 x + 2/3 cao
(b) Answer: y = 4/3 x + 2/3

Question 9

M1 set tangent gradient -x/y = 2 so x = -2y oe
M1 substitute: (-2y)² + y² = 25 so 5y² = 25 so y² = 5 oe
A1 points (-2 √5, √5) and (2 √5, -√5) cao
Answer: (-2 √5, √5) and (2 √5, -√5)

Question 10

M1 find tangent equation: slope -4/3 and intercept c with 3 = -4/3*4 + c so c = 25/3 oe
M1 find x-intercept A: 0 = -4/3 x + 25/3 so x = 25/4 and y-intercept B: (0, 25/3) oe
M1 area formula 1/2 * x_intercept * y_intercept oe
A1 area = 1/2 * 25/4 * 25/3 = 625/24 cao
Answer: 625/24

Question 11

M1 write tangent at (x0,y0) as x x0 + y y0 = 50 and substitute (10,0): 10 x0 = 50 oe
M1 so x0 = 5 and use circle to find y0² = 50 - 25 = 25 oe
M1 find y0 = ±5 oe
A1 points (5,5) and (5,-5) cao
Answer: (5, 5) and (5, -5)

Question 12

(a) M1 find gradient of L: 3x - 4y + 10 = 0 gives y = 3/4 x + 5/2 so gradient 3/4; perpendicular gradient -4/3 oe
(a) M1 set tangent gradient -x0/y0 = -4/3 so x0/y0 = 4/3 so x0 = 4k, y0 = 3k oe
(a) M1 substitute into circle: (4k)² + (3k)² = 169 so 25k² = 169 so k = ±13/5 oe
(a) A1 points (52/5, 39/5) and (-52/5, -39/5) cao
(a) Answer: (52/5, 39/5) and (-52/5, -39/5)
(b) M1 tangent gradient is -4/3 so form y = -4/3 x + c and substitute (52/5,39/5) to find c: 39/5 = -4/3 * 52/5 + c oe
(b) M1 compute c: multiply by 15 => 117 = -208 + 15c so 15c = 325 so c = 65/3 oe
(b) M1 other tangent through (-52/5,-39/5) gives c = -65/3 by symmetry oe
(b) A1 equations y = -4/3 x + 65/3 and y = -4/3 x - 65/3 cao
(b) Answer: y = -4/3 x + 65/3 and y = -4/3 x - 65/3

Perpendicular Lines and the equation of a tangent: Fluency and Exam Drill - Worksheets, Questions and Revision