Answer:
a is your answer
Step-by-step explanation:
;)
I have so much workkk but yea i’m good how are you
I: 12x-5y=0
II:(x+12)^2+(y-5)^2=169
with I:
12x=5y
x=(5/12)y
-> substitute x in II:
((5/12)y+12)^2+(y-5)^2=169
(25/144)y^2+10y+144+y^2-10y+25=169
(25/144)y^2+y^2+10y-10y+144+25=169
(25/144)y^2+y^2+144+25=169
(25/144)y^2+y^2+169=169
(25/144)y^2+y^2=0
y^2=0
y=0
insert into I:
12x=0
x=0
-> only intersection is at (0,0) = option B
the system has no solution.
Option C is correct.
Step-by-step explanation:
We need to solve the system of equations by substitution

Putting value of x from eq(2) into eq(1)

As as 0≠4is not true, we cannot find the value of y so the system has no solution.
Option C is correct.
Keywords: System of equations
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