Question

1. Find k if the following pairs of circles are orthogonal
() + y2 + 2by-k=0, x2 + y2 + 2ax +8=0.
(1) 2 + y2 - 6x – 8y + 12=0, x+7-4x+60 +6=0 /
(ii) x2 + y2 - 5x – 14y – 34 = 0, ** + y2 + 2x + 4y + k=0
(iv) x2 + y2 + 4x + 8 = 0, x2 + y2 - 16y +k=0
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Answer 1

Hii mate!!!

options are incomplete..

by the way ur ans is

(1) 2 + y2 - 6x – 8y + 12=0, x+7-4x+60 +6=0

HOPE IT HELPS..☺☺❤❤

Answer 2

Orthogonal Circles are those intersecting circles, the tangents to the point of intersection are at right angles to each other. The general formula for a circle is given by:

x²+y²+ 2gx + 2fy +c = 0

And if two circles are orhtogonal, they must satisfy the equation,

2$g_{1}$$g_{2}$ + 2$f_{1}$$f_{2}$ = $c_{1}$ + $c_{2}$

Therefore, to find k in the given equations;

(a) x²+y² + 2by - k = 0 and x²+y² + 2ax + 8 =0

∴ $g_{1}$ = 0; $g_{2}$ = 2a; $f_{1}$ = 2b; $f_{2}$ = 0; $c_{1}$ = -k and $c_{2}$ = 8

∴ 2$g_{1}$$g_{2}$ + 2$f_{1}$$f_{2}$ = $c_{1}$ + $c_{2}$

∴ 2×0×2a + 2×2b×0 = -k+8

∴ k = 8

(b) x²+y² - 6x - 5y + 12  0 and x²+7y² - 4x + ky + 6 = 0

∴ $g_{1}$ = -6; $g_{2}$ = -4 ; $f_{1}$ = -8; $f_{2}$ = k; $c_{1}$ = 12 and $c_{2}$ = 6

∴ 2$g_{1}$$g_{2}$ + 2$f_{1}$$f_{2}$ = $c_{1}$ + $c_{2}$

∴ k = 3

(c) x²+y² - 5x - 14y - 34 = 0 and x²+y² + 2x + 4y + k = 0

∴ $g_{1}$ = -5; $g_{2}$ = 2 ; $f_{1}$ = -14; $f_{2}$ = 4; $c_{1}$ = -34 and $c_{2}$ = k

∴ 2$g_{1}$$g_{2}$ + 2$f_{1}$$f_{2}$ = $c_{1}$ + $c_{2}$

∴ 2×-5×2 + 2×-14×4 = -34+k

∴ k = -98

(d) x²+y² + 4x + 8 = 0 and x²+y² - 16y + k = 0

∴ $g_{1}$ = 4; $g_{2}$ = 0; $f_{1}$ = 0; $f_{2}$ = -16; $c_{1}$ = 8 and $c_{2}$ = k

∴ 2$g_{1}$$g_{2}$ + 2$f_{1}$$f_{2}$ = $c_{1}$ + $c_{2}$

∴ 2×4×0 + 2×0×-16 = 8 + k

∴ k = -8