Question

encircle the linear equation in the following p+3=5​

Answer 1

(x−q)2+(y−p)2=r2(0)

A possible very elementary way is to use this formula thrice, one for each point. Since the circle passes through the point (5,10), it satisfies (0), i.e.

(5−q)2+(10−p)2=r2(1)

Similarly for the second point (−5,0):

(−5−q)2+(0−p)2=r2,(2)

and for (9,−6):

(9−q)2+(−6−p)2=r2.(3)

We thus have the following system of three simultaneous equations and in the three unknowns p,q,r:

⎧⎩⎨⎪⎪(5−q)2+(10−p)2=r2(−5−q)2+p2=r2(9−q)2+(6+p)2=r2(4)

To solve it, we can start by subtracting the second equation from the first

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪(5−q)2+(10−p)2−(5+q)2−p2=0(5+q)2+p2=r2(9−q)2+(6+p)2=r2

Expanding now the left hand side of the first equation we get a linear equation

⎧⎩⎨⎪⎪100−20q−20p=0(5+q)2+p2=r2(9−q)2+(6+p)2=r2

Solving the first equation for q and substituting in the other equations, we get

⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪q=5−p(10−p)2+p2−(4+p)2−(6+p)2=0(4+p)2+(6+p)2=r2

If we simplify the second equation, it becomes a linear equation in p only

⎧⎩⎨q=5−p48−40p=0(4+p)2+(6+p)2=r2

We have reduced our quadratic system (4) to two linear equations plus the equation for r2. From the second equation we find p=6/5, which we substitute in the first and in the third equations to find q=19/5 and r2=1972/25, i.e

⎧⎩⎨⎪⎪⎪⎪q=5−65=195p=65r2=(4+65)2+(6+65)2=197225.(5)

So the equation of the circle is

(x−195)2+(y−65)2=197225.

Answer 2

Answer:

p=2

Step-by-step explanation:

$p + 3 = 5 \\ \\ p = 5 - 3 \\ \\ p = 2$