Is p < 5? (1) In the coordinate plane, (3,p) lies inside the circle

We are to determine if p<5

Statement 1: In the coordinate plane, (3,p) lies inside the circle whose equation is x^2+y^2=16
The equation of a circle is (x-a)^2+(y-b)^2=r^2, where (a,b) is the center of the circle and r is the radius of the circle.
So given that the equation within which the point (3,p) lies is x^2+y^2=4^2, then the center of the circle is the origin (0,0) and r=4
And if p lies within the circle we can conclude that p<5 considering the maximum value of y-coordinate of any point on the above circle is 4.
Statement 1 is sufficient.

Statement 2: In the coordinate plane, the point (p,3) lies on the line 5x−3y=8
Basically, we just need to substitute the point (p,3) into the equation of the line 5x-3y=8 and solve for p
5p-9=8
p=17/5 = 3.4
Clearly p<5, hence statement 2 is also sufficient.

The answer is D.

Is p < 5?

(Statement1): In the coordinate plane, \((3,p)\) lies inside the circle whose equation is \(x^{2}+y^{2}=16\)
\(x^{2}+y^{2}=16=R^{2}\) --> \(R =4\)
The center of the circle is in the origin of the coordinate plane. \((0,0)\)
--> \(p\) must be \(-4 < p < 4\)
Sufficient

(Statement2): In the coordinate plane, the point \((p,3)\) lies on the line \(5x−3y=8\)
--> \(5p -3*3 =8\) --> \(p= \frac{17}{5}\)
Sufficient

Answer (D).

(1) sufic

x^2+y^2=16 is circle at origin and r^2=16, r=4
since r=4, and p is inside the circle, then p<5

(2) sufic

5x-3y=8…(3,p)…5x-3(3)=8…5x=17…x=p=17/5=3.4

Ans (D)

Is p < 5?

(1) In the coordinate plane, (3,p) lies inside the circle whose equation is \(x^2+y^2 = 16\)
3^2 + p^2 < 16 (since point lies inside of the circle with radius 4 units)
p^2 < 7
p < \(\sqrt{7}\) or p > -\(\sqrt{7}\)
Both cases less than 5

SUFFICIENT.

(2) In the coordinate plane, the point (p,3) lies on the line 5x−3y=8
5*p - 3*3 = 8
p = 3.4

SUFFICIENT.

Answer D.

(1) In the coordinate plane, (3,p) lies inside the circle whose equation is x^2+y^2=4^2....... radius of the circle is 4
So the coordinates of points inside the circle will b < 4
So p<4
Sufficient

(2) In the coordinate plane, the point (p,3) lies on the line 5x-3y=8
5p-3(3)=8
p=3.4
Sufficient

OA:D

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Ans. D

(1)
If (3,p) lies inside the circle whose equation is x2+y2=16x2+y2=16 ....... the 9 + p^2 <16 ------>> -sqrt(7) < p sqrt(7) ...... Hence Sufficient

(2) In the coordinate plane, the point (p,3) lies on the line 5x−3y=85x−3y=8 ........ . putting the value gives ------------> p = 5.6 .......... Hence Sufficient