CAT 2025 Slot 2 QA Question 1

Alright, so I've got this geometry problem here from the CAT exam. Let me try to visualize it first. There's a circle with center O, and a point P outside the circle. From P, two tangents are drawn, touching the circle at points Q and R. So, PQ and PR are both tangents from P to the circle.

Now, points A and B are on PQ and PR respectively, such that AB is also a tangent to the same circle. The angle AOB is given as 50 degrees, and We find the measure of angle APB.

Hmm, okay. Let me recall some properties of tangents to circles. the lengths of tangents from an external point to a circle are equal. So, PQ equals PR. Also, the line from the external point to the center of the circle bisects the angle between the two tangents. So, OP bisects angle QPR.

But in this problem, AB is another tangent. So, A is on PQ and B is on PR, and AB touches the circle at some point, say S. Since AB is a tangent, the radius OS is perpendicular to AB at point S.

I think I need to relate angle AOB with angle APB. Let me try to draw this in my mind. Points A and B are on PQ and PR, so triangle APB is formed. The angle at P is angle APB, which We find.

Given that angle AOB is 50 degrees, I wonder how that relates to angle APB. Maybe I can find some relationship between these angles using properties of circles and tangents.

Let me recall that the angle between two tangents from an external point is equal to twice the angle at the center subtended by the chord joining the points of contact. Wait, is that right? If I have two tangents from P, then the angle at P is equal to the angle subtended by QR at the center, but actually, it's supplementary to that angle. Because the angle between the tangents is equal to 180 degrees minus the central angle.

Wait, no, more accurately, the angle between the two tangents is equal to the angle subtended by the chord QR at the center. But actually, it's the other way around. The angle at the center is twice the angle at the external point. Hmm, I might be mixing things up.

We have the formula. The angle between two tangents drawn from an external point P is equal to twice the angle at the center O subtended by the chord QR. So, angle QPR = 2 angle QOR. But wait, angle QOR is the central angle, which is twice the angle at P. So, angle QPR = 2 angle QOR. Wait, no, that can't be because if angle QOR is, say, 60 degrees, then angle QPR would be 60 degrees as well, but that doesn't make sense because the angle at the center should be larger.

Wait, actually, the angle between the tangents is equal to the difference between 180 degrees and the central angle. So, angle QPR = 180 - angle QOR. Hmm, I'm getting confused here. Maybe I should look up the exact formula.

Wait, no, I think the correct formula is that the angle between the two tangents from an external point is equal to the angle subtended by the chord QR at the center. Wait, no, that's not right either. Let me think again.

the angle between the tangents is supplementary to the central angle. So, angle QPR + angle QOR = 180 degrees. So, angle QPR = 180 - angle QOR. That makes sense because if the central angle is 120 degrees, the angle at P would be 60 degrees, which is supplementary.

But in this problem, we're dealing with angle AOB, which is 50 degrees. So, angle AOB is the central angle subtended by chord AB. Since AB is a tangent, wait, no, AB is a tangent, so the chord would be the point where AB touches the circle, say S. So, angle AOB is the angle at the center between points A and B, but wait, A and B are points on the tangents, not on the circle.

Wait, no, AB is a tangent, so it touches the circle at one point, say S. So, the chord would be the point S, but that doesn't make sense. Wait, no, AB is a tangent, so it's just a single point of contact. So, angle AOB is the angle between the lines OA and OB, where A and B are points on the tangents PQ and PR.

Wait, but A and B are on the tangents, so OA and OB are not radii. Hmm, this is getting confusing. Maybe I should try to draw a diagram.

Let me try to sketch this mentally. There's a circle with center O. Point P is outside, and PQ and PR are tangents touching at Q and R. Points A and B are on PQ and PR respectively, such that AB is a tangent. So, AB touches the circle at some point S.

Now, OA and OB are lines from the center to points A and B on the tangents. The angle between OA and OB is given as 50 degrees. I need to relate this angle to angle APB.

Hmm, maybe I can use the property that the angle between two lines from the center to points on the tangents is related to the angle at P.

Wait, let me recall that the angle between the tangents from P is equal to the angle subtended by QR at the center. So, angle QPR = angle QOR. Wait, no, that's not right. I think it's the other way around. The angle at P is equal to half the difference of the intercepted arcs. Wait, that's for angles formed outside the circle.

Wait, yes, the formula is that the measure of an angle formed outside the circle is half the difference of the measures of the intercepted arcs. So, angle QPR = 1/2 (arc QR - arc QS), but I'm not sure.

Wait, no, more accurately, the angle between two tangents drawn from an external point is equal to half the difference of the measures of the intercepted arcs. So, angle QPR = 1/2 (measure of major arc QR - measure of minor arc QR).

But in this case, since AB is another tangent, maybe I can apply a similar concept.

Alternatively, maybe I can consider triangle APB and triangle AOB.

Wait, let me think about the properties of tangents. Since AB is a tangent, the radius at the point of contact is perpendicular to AB. So, if AB touches the circle at S, then OS is perpendicular to AB.

But I don't know where S is exactly. Maybe I can use some properties of similar triangles or something.

Wait, another approach: Since PQ and PR are tangents from P, and AB is another tangent, maybe I can use