Find the equation of the circle having centre at (3,2) and touching the line 4 + 3 − 8 = 0
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Answer:
Correct option is D)
x+y=8 is a tangent to a circle having center of (3,3)
then r= perpendicular distance of (3,3) from
x+y−8=0
Perpendicular distance of centre (x
1
,y
1
) from line ax+by+c=0 is
∣
∣
∣
∣
∣
a
2
+b
2
ax
1
+by
1
+c
∣
∣
∣
∣
∣
So, r=
∣
∣
∣
∣
∣
2
3+3−8
∣
∣
∣
∣
∣
=
2
2
=
2
The equation of the circle with centre (h,k) and the radius a is (x−h)
2
+(y−k)
2
=a
2
So, equation of circle is
(x−3)
2
+(y−3)
2
=(
2
)
2
x
2
+y
2
−6x−6y+18−2=0
⇒x
2
+y
2
−6x−6y+16=0
Step-by-step explanation:
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