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sherxon
Joined: 28 May 2021
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Location: Uzbekistan
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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:
65%
(hard)
Question Stats:
59% (02:42) correct
41%
(02:55)
wrong
based on 66
sessions
History
Date
Time
Result
Not Attempted Yet
The radius of the circle is 10 cm. The area of the triangle OAB is 600 \(cm^2\). OA>OB. Find the coordinate of the point P.
A. (15:20)
B. (12:18)
C. (16:18)
D. (16:20)
E. (15:18)
Attachment:
triangle.jpg [ 7.6 KiB | Viewed 2896 times ]
IanStewart
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Last visit: 10 Sep 2026
Posts: 4,146
Location: United States
GMAT 1: 780 Q51 V47
Expert reply
GMAT 1: 780 Q51 V47
Posts: 4,146
The center of the circle is at (10, 10). Every point on the circle must be a distance of 10 from the center. Going from the center to answer choice C, (16, 18), we'd go right 6 units and up 8 units, so we'd get a 6-8-10 triangle, and answer choice C is on the circle. No other answer choice is, so that answer must be correct.
I didn't try to solve the question properly (i.e. without answer choices), but at a glance, it doesn't look like it would be pleasant, algebraically speaking.
_________________
I didn't try to solve the question properly (i.e. without answer choices), but at a glance, it doesn't look like it would be pleasant, algebraically speaking.
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sherxon
Joined: 28 May 2021
Last visit: 22 Aug 2026
Posts: 130
Location: Uzbekistan
Concentration: Strategy, General Management
Schools: Carlson (A$$$) Kenan-Flagler '25 (A$$$$) Goizueta '25 (A$$$) Olin '25 (A$$$$) Fuqua '25 (M$$$$) Marshall '25 (A$$$$)
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q50 V46 (Online)
GPA: 4
Products:
Schools: Carlson (A$$$) Kenan-Flagler '25 (A$$$$) Goizueta '25 (A$$$) Olin '25 (A$$$$) Fuqua '25 (M$$$$) Marshall '25 (A$$$$)
GMAT 2: 770 Q50 V46 (Online)
Posts: 130
IanStewart
You solved it the gmat way)
Posted from my mobile device
Since the graph is at right angle the triangle should follow a triplets. Hence its length ate in ratio 3:4:5 3 and 4 being the base and height (3*4=12) we can confirm this by getting base * height from triangle area.
So now we know height is 30 and base is 40.
Max height of circle on y axis is 20(diameter) and since the tangent is at angle height will be slightly less than 20
Same analogy at x axis and we can deduce it is closer to center hence it should be slightly more than 10. So my guess is
12 :18 (B)
Posted from my mobile device
So now we know height is 30 and base is 40.
Max height of circle on y axis is 20(diameter) and since the tangent is at angle height will be slightly less than 20
Same analogy at x axis and we can deduce it is closer to center hence it should be slightly more than 10. So my guess is
12 :18 (B)
Posted from my mobile device
Thelegend2631
Joined: 04 May 2020
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raju1212
You were right until first part.
You might want to refer to IanStewart solution above.
Centre will be at 10,10
Distance from this point to any point given in the options would be 10
Only C satisfies the condition
Or else
If things get trickier with different options
Since we know area and incentre.
\(Semi perimeter = Area/Incentre Radius\)
Simple
S= 600/10 = 60
Sum of sides = 120
So we know this could be 30,40,50 triplet.
If you were given multiple options that satisfies distance is 10 units.
Since we know intercepts of x and y axis by knowing the lengths of the legs
The hypotenuse will form
\(X/40 + Y/30 = 1\)
So only one such option or one such coordinate will pass through the above equation or satisfy the above condition
Here option C
16/40 + 18/ 30 = 1
1 = 1
Hence c
IanStewart
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GMAT 1: 780 Q51 V47
Posts: 4,146
CEO2021
If you draw the picture, but draw line AB with a slope of -1, then since AB is a tangent line to the circle, it's perpendicular to the radius, so the radius that meets point P would then have a slope of 1 (the negative reciprocal of -1). But that radius also passes through the center of the circle, (10, 10), so the radius line would have the equation y = x, and the point P would then have equal x and y coordinates (the point would be (10 + 5√2, 10 + 5√2).) That's where the point would be if the slope of AB was -1, which is not the situation we have. If you then imagine moving the points A and B, moving B up and A to the left, so that line AB falls more steeply (but still touches the circle at one point), you'll see that the point P moves down and to the right, and its x-coordinate grows while its y-coordinate shrinks. If instead you move A to the right, and B down, so the slope of AB gets closer to zero, then point P moves up and to the left. So its x-coordinate gets smaller and its y-coordinate gets larger. That's what we're doing in this question, because OA > OB, so the x-coordinate of P must be less than the y-coordinate of P, and if we saw the options (18, 16) and (16, 18), we'd know (16, 18) was right. The answer (18, 16) would be right if instead OB > OA.
If instead you're asking how to solve without answer choices, then, if you're only using GMAT-level math, I don't think you can, or at least not without doing an enormous amount of work (unless you just make a lucky guess at the dimensions of the triangle). The solution hD13 outlined above looks like the fastest one to me, but it uses facts you'd never need on the test. When you inscribe a circle in a triangle, there is a relationship between the circle's radius and the perimeter and area of the triangle: (half the perimeter)*(radius) = Area. You would never need to know that on the GMAT, but using it here, we find the perimeter is 120. If the area is 600 and the perimeter is 120, there's only one triangle we can have (if you don't care which side is shortest), and with the perimeter and area of the triangle, you can either solve algebraically or see by inspection that we have a 30-40-50 triangle. Then there are a few options to finish the problem - for example, since we now have two points on the line, we can produce its equation. We also can produce an equation for the circle (though you really never need to do that on the GMAT for circles that are not centred at the origin) and just solve algebraically for their intersection point. Or with the equation of the line, you can then easily distinguish between answer choices if several answers lie on the circle.
Real GMAT questions aren't set up this way though, where the only convenient way to solve is by eliminating the answers that don't make sense (that aren't on the circle). In a real GMAT question, there will always be some practical two-minute way to arrive at the answer if you 'do the math'.
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Thelegend2631
Joined: 04 May 2020
Last visit: 18 Nov 2025
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Status:What goes round comes around, so does Kudos.
Location: India
GPA: 3
WE:Business Development (Retail Banking)
Posts: 371
CEO2021
Since we know area and incentre.
Semiperimeter=Area/IncentreRadiusSemiperimeter=Area/IncentreRadius
Simple
S= 600/10 = 60
Sum of sides = 120
So we know this could be 30,40,50 triplet.
If you were given multiple options that satisfies distance is 10 units.
Since we know intercepts of x and y axis by knowing the lengths of the legs
The hypotenuse will form
X/40+Y/30=1X/40+Y/30=1
So only one such option or one such coordinate will pass through the above equation or satisfy the above condition
Here option C
16/40 + 18/ 30 = 1
1 = 1
Hence c
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