Creating a circle equation that has a specific slope at a specific point

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basically I need to create a circle that will connect with a linear equation at a point so that it is both a continuous and smooth transition, meaning at that point the circle needs to have both the same derivative and y value as the linear equation. Right now I'm playing around with the equation (ax + b)^2 + (ax + c)^2 = 1 where a b and c are all constants that I'm trying to find out so that when x = 1, y = 4 and the derivative is also 4. I know how to get a line to fit a circle (a tangent) so that its smooth and continuous but I don't know how to create a circle to fit a line, help!

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All the tangents circles with center lying on the perpendicular at the tangent line at that point are solutions.

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