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secondary 4 | A Maths
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I dont know how to do part(iv), i tried using simultaneous equations but theres a lot of steps and doesnt seem to be worth 4marks
Date Posted: 3 years ago
Views: 233
AQ is a radius of C1 and QR is part of the tangent to C1.
Circle property : tangent is perpendicular to radius. So AQ ∟ QR and ∠AQR = 90° (right angle)
① So△AQR is a right-angled triangle.
② Since A, Q and R are points on C2, ∠AQR is a right angle in a semicircle (recall circle properties)
③ AR is the hypotenuse of △AQR. Using the inference in ②, it can deduced that AR is a diameter of C2.
Coordinates of the centre of C2, B
= coordinates of the midpoint of AR
= ( (-2 + 7/3)/2 , (-3 + 0)/2 )
= (1/6, -3/2)
You will be able to find the radius of C2 from here.
Circle property : tangent is perpendicular to radius. So AQ ∟ QR and ∠AQR = 90° (right angle)
① So△AQR is a right-angled triangle.
② Since A, Q and R are points on C2, ∠AQR is a right angle in a semicircle (recall circle properties)
③ AR is the hypotenuse of △AQR. Using the inference in ②, it can deduced that AR is a diameter of C2.
Coordinates of the centre of C2, B
= coordinates of the midpoint of AR
= ( (-2 + 7/3)/2 , (-3 + 0)/2 )
= (1/6, -3/2)
You will be able to find the radius of C2 from here.
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