Point W = (5, 3). Circle J has a center at point W and radi

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Mike@Magoosh: GMAT Instructor; Posts: 768; Joined: Wed Dec 28, 2011 4:18 pmFollowed by: 140 members; Location: Berkeley, CA; Thanked: 387 times

by Mike@Magoosh » Fri Apr 25, 2014 12:46 pm

Point W = (5, 3). Circle J has a center at point W and radius of r = 5. This circle intersects the y-axis at one intercept and the x-axis at two intercepts. What is the area of the triangle formed by these three intercepts?
(A) 7.5
(B) 12
(C) 15
(D) 24
(E) 30

This question is medium difficulty. If you would like to practice on a set of medium to very hard problems on Coordinate Geometry, as well as see the OE of this question, see:
https://magoosh.com/gmat/2014/challengin ... questions/

OA = [spoiler](B)[/spoiler]

Mike

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Source: Beat The GMAT — Problem Solving | Fri Apr 25, 2014 12:46 pm

Uzbekguy: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Mon Nov 16, 2009 7:26 pm

by Uzbekguy » Tue Aug 19, 2014 10:19 am

The area=(3x9)/2-(3*1)/2=(3(9-1))/2=12

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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

by GMATinsight » Tue Aug 19, 2014 10:42 am

Mike@Magoosh wrote:Point W = (5, 3). Circle J has a center at point W and radius of r = 5. This circle intersects the y-axis at one intercept and the x-axis at two intercepts. What is the area of the triangle formed by these three intercepts?
(A) 7.5
(B) 12
(C) 15
(D) 24
(E) 30

This question is medium difficulty. If you would like to practice on a set of medium to very hard problems on Coordinate Geometry, as well as see the OE of this question, see:
https://magoosh.com/gmat/2014/challengin ... questions/

OA = [spoiler](B)[/spoiler]

Mike

Equation of Circle with Radius 5 units and center at (h,k) => (X-h)^2 + (Y-k)^2 = 5^2
i.e. (X-5)^2 + (Y-3)^2 = 25

This circle intersects the y-axis at one intercept which mean the Circle is Tangent to the Y-Axis

For point of Intersection at Y-Axis, X = 0

Substitute it in the Equation of Circle
i.e. X^2 + Y^2 = 25 ===> i.e. (-5)^2 + (Y-3)^2 = 25
i.e. Y = 3

For X-Intercepts of the Circle, Y = 0
i.e. (X-5)^2 + (Y-3)^2 = 25 ===> (X-5)^2 + (0-3)^2 = 25 ===> (X-5)^2 = 16
i.e. X-5 = +4 ===> i.e. X = 9, 1

Therefore Point of Intersections are (9,0) (1,0) and (0,3)
This is a triangle with Base as Distance between (9,0) and (1,0) = 8
and Height as Y-coordinate of point (0,3) = 3
Therefore Area of Triangle = (1/2) x 8 x 3 = 12

Answer: Option D

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