Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
aatish agrawal Grade: Upto college level
`        find ind the equation of the circle which pass through the focus of the parabola x^2=4ay and touches it at point (6,9)`
6 years ago

gOlU g3n|[0]uS
42 Points
```
```
6 years ago
510 Points
```										x2 = 4ay    ...........1
since circle touches the parabola at (6,9) so this point will satisfy the equation of both the curves
so , from eq 1
36 = 36a
a=1 , so eq 1 becomes
x2 = 4y
now differentiating above eq
2x = y1 = m (slope of tangent)
at (6,9) , m = 3        ..........2
let (h,k) be the center then slope of line passing through center of circle & the point of contact is given by
m1 = (k-9)/(h-6)        .........3
mm1 = -1   or
(k-9)/(h-6) . 3 = -1
h + 3k = 33                      ................4
now focus of parabola is (0,1) ....
eq of circle is x2+y2 +2gx+2fy+c = 0                              (g=-h , f=-k)
so eq of circle : x2+y2-2hx-2ky+c
this circle passes through (0,1) & (6,9) so its eq will satisfy these points therefore
-2k+1+c = 0            ...................5
-12h-18k+117+c = 0             ....................6
from 5 & 6
3h+4k=29          ............7
solving 4 & 7 we get
(h,k) = (-9,14)
c = 27
now required eq of circle is
x2 + y2 18x - 28y + 27 = 0
```
6 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Discuss with colleagues and IITians

View all Questions »
• Complete AIPMT/AIIMS Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details